Chapter 3 Kinematics
The area under a velocity vs time graph gives the ___________
displacement
Gravitational acceleration value
9.8 m/s^2 or 10 m/s^2
Express a mass of 10 kg in grams. A. 10^4 g B. 10^-4 g C. 10^2 g D. 10^-2 g
A
On Earth, a tennis player can hit a ball normally, causing the ball to travel on a path that is symmetrical parabola. A tennis player can also hit a tennis ball with a "slice", which causes the ball to spin and deviate to one side of its normal path. What is the best explanation for this deviation? A. There is an additional acceleration on the ball B. The spin on the ball caused the acceleration from gravity to change its direction C. The spin on the ball used energy to the ball could not travel in a straight line D. The gravitational field was not uniform
A
The work done in pushing an object of mass m (in kg) from rest to speed v (in m/s) is given by W = mv2/2. Find the SI unit for work. A. kg·m2/s2 B. kg·m/s C. kg2·m/s2 D. kg2·m2/s2
A
A racehorse makes one lap around a 500-meter track in a time of 25 seconds. What was the racehorse's average speed? A. 0 m/s B. 5 m/s C. 20 m/s D. 10 m/s
C
A car accelerates uniformly in a straight line from 0 to 60 mi/h in 6 seconds. What is the magnitude of the acceleration? A. 10 mi·s-2 B. 10 mi·h-2 C. 10 mi·h-1 D. 10 mi·h-1·s-1
D
instantaneous velocity
the speed and direction of an object at a particular instant
Kinematics equation (no displacement)
v = v0 + at
cos 90
0
sin 0
0
cos 0
1
sin 90
1
Acceleration can change when:
1. An object speeds up 2. An object slows down 3. An object changes direction
cos 60
1/2
sin 30
1/2
On the way down, an object gains ______ m/s^2 every second
10
On the way up, an object loses __________ m/s^2 every second
10
picometer
10^-12 m
centimeter
10^-2 m
millimeter
10^-3 m
micrometer
10^-6 m
Nanometer
10^-9 m
kilometer
10^3 m
megameter
10^6 m
Gigameter
10^9 m
A particle's speed at time t is given by v = kt2, where k is some constant. What are the dimensions of k? (L = length, T = time) A. LT-3 B. LT-2 C. LT-1 D. LT
A;
Jose asks Jill to slide the salad bowl (m = 2 kg) across the table. Jill's initial push gives it an initial speed of 4 m/s, and it comes to a halt directly in front of Jose after traveling 2 meters. What was the magnitude of the kinetic frictional force slowing the bowl during its slide? A. 8 N B. 4 N C. 2 N D. 0.8 N
A; v0 = 4 m/s, v = 0 m/2, x = 2m, a = ? v^2 = v^2 + 2ad 0 = 16 + 4a -16 = 4a a = -4m/s^2 Fnet = ma Fnet = 2(-4) Fnet = -8 N
An arrow is shot straight up, and it eventually falls straight back down. Ignoring air resistance, which one of the following statements concerning the acceleration a of the arrow is correct? A. a always points down. B. a always points in the direction of the velocity. C. a always points up. D. a always opposes the velocity.
A; "a always points down" is correct concerning the acceleration of an arrow that is shot straight up and eventually falls straight back down. Once released, the only acceleration acting on the arrow (ignoring air resistance) is that due to gravity. On Earth, the acceleration due to gravity (the vector g) will always point down, toward the center of the Earth. Though the arrow's velocity changes directions, the acceleration due to gravity on Earth remains constant.
A particle moving along the x axis passes through the point x = 0 (in the -x direction) at a particular instant. If it experiences a constant acceleration of -2 m/s2, where could the object be three seconds later? A. x = -12 m B. x = -3 m C. x = -6 m D. x = -9 m
A; A particle moving along the x axis passes through the point x = 0 (in the -x direction) at a particular instant. If it experiences a constant acceleration of -2 m/s2, the object could be at x = -12 m three seconds later. Using Big Five #3, with ∆x = x (since x0 = 0): Since v0 is negative (because we're told that the object is traveling in the -x direction at the beginning of the time interval), x must be less than -9. The only possibility among the given choices is x = -12 m.
An arrow is projected straight up with an initial velocity of v0 m/s. If g denotes the magnitude of the gravitational acceleration (in m/s2) and air resistance is ignored, what is the arrow's maximum height (in m)? A. v02/2g B. 2v02/g C. v0/2g D. v0/g
A; An arrow is projected straight up with an initial velocity of v0 m/s. If g denotes the magnitude of the gravitational acceleration (in m/s2) and air resistance is ignored, the arrow's maximum height is v02/2g. One solution uses Big Five #5. All we have to realize is that at the top of its (purely vertical) flight, its velocity v = 0. If we call "up" the positive direction, then the acceleration a = -g is negative, and we find
An object which is accelerating must be: A. changing its velocity. B. increasing its speed. C. changing its direction. D. traveling in a straight line.
A; An object which is accelerating must be changing its velocity. By definition, acceleration implies a change in velocity. ("Changing its direction" is wrong because an object can accelerate without changing its direction [just drive in a straight line and step on the gas]. "Traveling in a straight line" and "increasing its speed" are wrong because an a turning object moving at constant speed is constantly accelerating.)
An ideal projectile launched from the ground at an angle of 45° up from the horizontal has a total flight time of T. Which of the following correctly expresses a relationship between the displacement d, velocity v, and acceleration a at T / 2 compared to time T (the instant before it hits the ground)? A. 2 × |dT/2| > |dT| B. |vT/2| = |vT| C. |aT/2| > |aT| D. v T/2 = 0
A; At half the total flight time, a projectile launched upward from even ground upon which it will eventually land will be at the apex of its flight. At this time, all of its velocity is in the horizontal direction (meaning it has slowed down, eliminating choice B, but is still moving, eliminating choice D). Its acceleration throughout the flight is a constant g downward, eliminating choice C. That only leaves answer choice A, which is correct: The magnitude of the displacement from the launch point to the apex is a diagonal line that is the hypotenuse of a right triangle with the max height of the projectile as one leg and half the range (R = |dT|) as the other. Twice this hypotenuse must therefore be greater than the total range dT.
A racehorse makes one lap around a 500-meter track in a time of 25 seconds. What was the racehorse's average velocity? A. 10 m/s B. 0 m/s C. 5 m/s D. 20 m/s
B
An object is moving along a one-dimensional path. At its turning point: A. the acceleration is zero. B. the instantaneous velocity is zero. C. all of these statements are true. D. the average velocity of the object is zero.
B
The density of aluminum is approximately 2700 kg per m3. What is this density in g/cm3? A. 27 g/cm3 B. 2.7 g/cm3 C. 2.7 × 10^4 g/cm3 D. 0.27 g/cm3
B
Which of the following expressions could be interpreted as velocity? A. 5 m to the north B. 5 m/s to the north C. 5 m/s D. 5 m
B
A car, originally traveling at 10 m/s, accelerates uniformly for 4 seconds at a rate of 2 m/s2 in the direction of its velocity. How far does it travel during this period? A. 80 m B. 48 m C. 56 m D. 72 m
C; A car, originally traveling at 10 m/s, accelerates uniformly for 4 seconds at a rate of 2 m/s2 in the direction of its velocity. During this period, it travels 56 m. Apply Big Five #3 (Note the absence of units in the calculation: for the sake of speed, it is fine to exclude units when you write out your math if you're using a known equation and all the units are already in S.I. form.): d = vot + 1/2 at^2 d = (10)(4) + (.5)(2)(16) d = 40 + 16 d = 56 m
A car, originally traveling at 10 m/s, accelerates uniformly for 4 seconds at a rate of 2 m/s2. During this period, it travels 56 m. What is the magnitude of the car's average velocity during the acceleration period? A. 18 m/s B. 12 m/s C. 14 m/s D. 28 m/s
C; A car, originally traveling at 10 m/s, accelerates uniformly for 4 seconds at a rate of 2 m/s2. During this period, it travels 56 m. The magnitude of the car's average velocity during the acceleration period was 14 m/s. Avg velocity = delta x / delta t Avg velocity = 56/4 = 14m
A ball is tossed straight up in the air off of a bridge with an initial speed of 5 m/s. If it hits the water below the bridge with a speed of 15 m/s, from what height above the water was it tossed? A. 5 m B. 10 m C. 15 m D. 30 m
B;
A car moving 20 m/s in the positive x-direction slams on the brakes and stops after skidding 20 m. What was its acceleration during this period? A. -20 m/s2 B. -10 m/s2 C. -5 m/s2 D. -2 m/s2
B;
The equation for the Reynolds number, which determines whether the flow of a fluid is laminar or turbulent, is R = ρvL / µ, where ρ is density, v is the flow speed, L is length, and µ is the dynamic viscosity of the fluid (measured in kg / m·s). What are the units of the Reynolds number? A. 1 / m B. unitless C. kg / m D. 1 / m·s
B;
Where in its path does a projectile in free fall near the surface of the earth experience the greatest acceleration? (Assume no air resistance) A. While it is descending. B. Acceleration is the same at all points in the path. C. While it is ascending. D. At its greatest height.
B;
In a crash simulation, a car traveling at x m/s can stop at a distance of d m with a maximum deceleration. If the car is traveling at 2x m/s, which of the following statements is/are true, assuming maximum deceleration? I. The stopping time is doubled II. The stopping distance is doubled III. The stopping distance is quadrupled. A. I and II only B. I and III only C. II only D. III only
B; v^2 = vo^2 + 2ad v = v0 + at
A body is undergoing uniformly accelerated motion over a period of time. Which of the following is true? A. The final velocity of the object is equal to its average velocity. B. The relationship between the final velocity of the object and its average velocity cannot be determined from the information given. C. The final velocity of the object is less than its average velocity. D. The final velocity of the object is greater than its average velocity.
B; A body is undergoing uniformly accelerated motion over a period of time. The following is then true: The relationship between the final velocity of the object and its average velocity cannot be determined from the information given. If the velocity during the period of time in question were constant in magnitude but varied in direction, then the statement "The final velocity of the object is greater than its average velocity" and the statement "The final velocity of the object is less than its average velocity" would be false (since the average velocity would equal the final velocity). And, if the initial and final velocities were different in magnitude, then the statement "The final velocity of the object is equal to its average velocity" would be false.
A projectile is launched horizontally from a raised platform. If air resistance is ignored, then as the projectile falls to the earth, the magnitude of the vertical component of the velocity of the projectile: A. decreases, while the horizontal component increases. B. increases, while the horizontal component remains constant. C. decreases, while the horizontal component remains constant. D. increases, while the horizontal component decreases.
B; A projectile is launched horizontally from a raised platform. If air resistance is ignored, then as the projectile falls to the earth, the magnitude of the vertical component of the velocity of the projectile increases, while the horizontal component remains constant. Since the projectile experiences no horizontal acceleration, there will be no change in the horizontal component of its velocity. This eliminates "increases, while the horizontal component decreases" and "decreases, while the horizontal component increases". On its way down, however, the projectile's vertical speed increases, since it experiences the downward acceleration due to earth's gravity.
A typical speck of dust has a mass of 700 ng. How many specks of dust would it take to make 1 kg? A. 7.0 × 10^10 B. 1.4 × 10^9 C. 7.0 × 10^7 D. 1.4 × 10^6
B; A typical speck of dust has a mass of 700 ng. It would take 1.4 × 109 specks of dust to make 1 kg. Divide the total desired mass by the mass of a single speck: n = 1 kg / 700 ng = 10^3 g / 7 x 10^-7 g = 1/7 x 10^10 = 1.4 x 10^9
A 2-kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon. If it returns to its starting point in 4 seconds, what is the acceleration due to gravity on the moon? A. 3.2 m/s2 B. 1.6 m/s2 C. 0.8 m/s2 D. 6.4 m/s2
B; If a 2-kg rock is thrown vertically upward at a speed of 3.2 m/s from the surface of the moon and it returns to its starting point in 4 seconds, the acceleration due to gravity on the moon is 1.6 m/s2. Since the round-trip time is 4 seconds, the time it takes the rock to reach the top of its path (where v= 0) is half this: 2 seconds. Then, applying the equation a = ∆v/∆t to the first half of the trip, we find a = (0 - 3.2)/2 = -1.6 m/s2, so the magnitude of the moon's gravitational acceleration is 1.6 m/s2.
In an experiment, Ball A is dropped from the top of a building on Earth, and Ball B of equal mass is dropped from the top of a building twice as high but on a planet that has half of Earth's gravitational pull. Ignoring air resistance, which of the following is true? A. The final velocity of Ball A upon reaching the ground is higher. B. The final velocities of the two balls upon reaching the ground are the same. C. The two balls reach the ground at the same time. D. The final velocity of Ball B upon reaching the ground is higher.
B; In an experiment, Ball A is dropped from the top of a building on Earth, and Ball B of equal mass is dropped from the top of a building twice as high but on a planet that has half of Earth's gravitational pull. Ignoring air resistance, it is true that the final velocities of the two balls upon reaching the ground are the same. The choice, "the two balls reach the ground at the same time" could be eliminated first, because if Ball B is experiencing half the acceleration but double the displacement of Ball A, Ball B must reach the ground after Ball A. Big Five #5 pinpoints which of the remaining choices is correct. The formula v2 = v02 + 2ad is applied. Since the balls start at rest, v0 is zero. Rearranging for v, the equation becomes . From the equation, if we double the displacement, 2d, but halve the acceleration, a / 2, the final velocity remains unaffected.
An airplane is flying east a distance of 2000 km at a velocity, v. Another airplane is flying the same distance at the same velocity, v, but a distance of 2000 km to the west. Both planes are subjected to a constant wind blowing east at a velocity of (1/3)v. Compared to the time for east bound airplane's trip, how many times longer is the west bound airplane's trip? A. 3 times longer B. 2 times longer C. 1/3 times longer D. 1/2 times longer
B; The time for each airplane's trip can be found using the formula t = d / v. For the east bound airplane's trip, the total velocity is the sum of the airplane's velocity and the wind's velocity, so veast= v + (1/3)v = (4/3)v. The time for the east bound airplane is teast = 2000 / ((4/3)v) = 6000 / (4v) = 1500 / v. For the west bound airplane's trip, the total velocity is the airplane's velocity minus the wind's velocity, so vwest = v - (1/3)v = (2/3)v. The time for the west bound airplane is twest = 2000 / ((2/3)v) = 6000 / (2v) = 3000 / v = 2(1500 / v) = 2(teast). The correct answer is "2 times longer." Alternatively, note that the ground speed of the west bound plane is half that of the east bound plane, so it takes twice as long to reach its destination.
Two projectiles are launched from the same point. Projectile A has the greater horizontal velocity, while Projectile B has the greater vertical velocity. Which projectile will travel the greater horizontal distance? A. Both projectiles will travel the same distance. B. Cannot be predicted from the information given. C. Projectile A, because distance traveled is determined by horizontal velocity. D. Projectile B, because it will be in the air for more time.
B; Two projectiles are launched from the same point. Projectile A has the greater horizontal velocity, while Projectile B has the greater vertical velocity. Which projectile will travel the greater horizontal distance cannot be predicted from the information given. The horizontal distance traveled by a projectile—a quantity called its range—is equal to its horizontal velocity, v0x, multiplied by its total flight time (assuming that the take-off and landing points are at the same horizontal level). The total flight time is equal to twice the time necessary to reach the top of its path, which is where its vertical velocity drops to zero. The equation ∆vy = g∆tgives ∆t = v0y/g as the time to reach the topmost point, so the total flight time is twice this: 2v0y/g. Thus, the projectile's range is R = 2v0xv0y/g. Since R depends on both v0x and v0y, there is insufficient information given here to decide which projectile will have the greater range.
A block with a mass of 5 kg rests on a horizontal table (µs = 0.3, µk = 0.2). A constant horizontal pushing force of 20 N is applied. How far does the block move in 2 seconds? A. 0 m B. 2 m C. 4 m D. 20 m
C
It is well known that the flash and the sound of thunder produced by a lightning bolt are not observed simultaneously. This is due to the fact that light waves travel so much faster than sound waves. Light travels so quickly that one can assume that lightning bolts occur at the same time one sees them. Given that sound propagates through air at a speed of 340 m/s, how far away is a lightning bolt if the delay between hearing and seeing it is 5 sec? A. 68 m B. 4250 m C. 1700 m D. 6120 m
C
The horizontal component of the initial velocity of a projectile is directly proportional to the: A. sine of the angle of elevation. B. angle of elevation. C. cosine of the angle of elevation. D. tangent of the angle of elevation.
C
Which one of the following formulas could give the pressure P [in kg/(m·s2)] at depth h (in m) below the surface of the ocean, where ρ (the density of seawater) has units of kg/m3, and g has units of m/s2? A. P = ρh/g B. P = ρg/h C. P = ρgh D. P = gh/ρ
C
A ball is thrown in a projectile motion trajectory with an initial velocity v at an angle theta above the ground. If the acceleration due to gravity is -g, which of the following is the correct expression of the time it takes for the ball to reach its highest point, y, from the ground? A. v^2 sin theta / g B. -v cos theta / g C. v sin theta / g D. v^2 cos theta / g
C;
A bubble is a glass of beer releases from rest at the bottom of the glass and tires at acceleration a to the surface in t seconds. If t > 2, how much farther does the bubble travel between times t = 1 and t = 2 s the it does between times t = 0s and t = 1s? A. 2a meters B. 3a/2 meters C. a meters D. a/2 meters
C;
It takes light from the sun 8 minutes to reach the earth. If light travels 3 × 10^8m/s, how far is the sun from the earth in meters? A. 2.4 × 10^9 m B. 1.44 × 10^10 m C. 1.44 × 10^11 m D. 2.4 × 10^11 m
C; (3 x 10^8 m/s) x (60s / 1 min) x 8 = 1.44 x 10^-11 m
An object is projected upward at a 30° angle with the horizontal with an initial speed of 60 m/s. How long will it take to reach the top of its trajectory? (Ignore air resistance.) A. 4.5 sec B. 1.5 sec C. 3.0 sec D. 6.0 sec
C; An object is projected upward at a 30° angle with the horizontal with an initial speed of 60 m/s. Ignoring air resistance, it will take 3.0 sec to reach the top of its trajectory. The top of the parabolic trajectory is characterized by the fact that at this point the vertical velocity vy is zero. Knowing initial and final velocities and acceleration, we use Big Five #2 to find time: v = vo + at 0 = vo(sin30) - 10t 0 = (60)(.5) - 10t -30 = -10t t = 3
An object travels along the x axis at a constant speed of 3 m/s in the -x direction. If the object is at x= 4 m at t = 0, where is it at time t = 4 s? A. x = -4 m B. x = -16 m C. x = -8 m D. x = -12 m
C; An object travels along the x axis at a constant speed of 3 m/s in the -x direction. If the object is at x= 4 m at t = 0, it is at x = -8 m at time t = 4 s. The object's velocity is a constant v = -3 m/s, so using Big Five #1 (with ), we find ∆x = v∆t = (-3 m/s)(4 s) = -12 m Thus, x - x0 = -12 m, so x = -8 m since x0 = 4 m.
Car #1 starts to accelerate from rest just as Car #2 passes it. If Car #2 maintains a constant velocity of 20 m/s, and Car #1 accelerates uniformly at 5/8 m/s2, how long will it take for Car #1 to overtake Car #2? A. 16 sec B. 32 sec C. 64 sec D. 8 sec
C; Car #1 starts to accelerate from rest just as Car #2 passes it. If Car #2 maintains a constant velocity of 20 m/s, and Car #1 accelerates uniformly at 5/8 m/s2, it will take Car #1 64 sec to overtake Car #2. Car #1 will catch up to Car #2 when the two cars have the same position. Therefore, we must find the positions of the cars and set them equal to each other. The position of Car #2 is easy to calculate since the speed is constant: Using distance equals rate times time, we find x2 = 20t. For Car #1 we use Big Five #3, with v0 = 0 and a = 5/8: we get x1 = (5/16)t2. Setting x1 and x2 equal to each other and solving for t, we find that t = 64 seconds. Another solution involves finding how long it takes for Car #1 to reach the speed of Car #2. Using v = at, we get 20 = (5/8)t, so t = 32 seconds. Since it takes that long just for Car #1 to reach Car #2's speed, it will take even more time to actually catch up to Car #2's position, so the only possibility among the given choices is 64 seconds.
In a series of experimental trials, a projectile is launched with a fixed speed, but with various angles of elevation. As the angle of elevation is increased from 0° to 90°, the vertical component of the initial velocity: A. decreases, while the horizontal component remains constant. B. increases, while the horizontal component remains constant. C. increases, while the horizontal component decreases. D. decreases, while the horizontal component increases.
C; In a series of experimental trials, a projectile is launched with a fixed speed, but with various angles of elevation. As the angle of elevation is increased from 0° to 90°, the vertical component of the initial velocity increases, while the horizontal component decreases. The horizontal component of the initial velocity is v0x = v0 cos θ, and the vertical component is v0y = v0 sin θ. Therefore, as θ increases from 0° to 90°, v0x decreases (because cos θ decreases) and v0yincreases (because sin θ increases).
An object accelerates uniformly from rest and moves a distance X meters in time T seconds. If the object accelerates uniformly from rest with the same acceleration, how far does it move in time 2T seconds? A. (1.4)X meters B. 2X meters C. 4X meters D. Impossible to determine with the given information.
C; The first sentence gives the variables d, v0, and t, and from this we could find either a or v. One option would be to solve for a and use that with the information given in the second sentence to find d for t = 2T. However, it is more efficient simply to note that with v0 = 0, d ∝ t2, since d = v0t + 0.5at2. Hence, if time doubles (with all other conditions being equal), distance must quadruple.
The magnitude of your displacement after walking 3 mi north, then 6 mi west, then 5 mi north is 10 mi. If it took 1 h (= 1 hour) to walk the 3-mi section and 1.5 h to walk each of the last two sections, what was the magnitude of your average velocity? A. 3 4/9 mi/h B. 3 1/2 mi/h C. 2 1/2 mi/h D. 2 mi/h
C; The magnitude of your displacement after walking 3 mi north, then 6 mi west, then 5 mi north is 10 mi. If it took 1 h (= 1 hour) to walk the 3-mi section and 1.5 h to walk each of the last two sections, the magnitude of your average velocity is 2 1/2 mi/h. Since the total travel time is 1 + 1.5 + 1.5 = 4 hours, the average velocity (net distance divided by time) is
The slope at a point on a velocity versus time graph represents what physical quantity? A. Displacement B. Instantaneous velocity C. Instantaneous acceleration D. Average acceleration
C; The slope of a curve is rise over run. In this case, rise (y) is velocity and run (x) is time, yielding velocity / time or acceleration. This eliminates choices A and B. If the slope is measured at a particular point instead of between two points, the quantity represented is instantaneous, not average, eliminating choice D.
A soccer ball is kicked at a speed of 30 m/s at an angle of 30° up from the horizontal. How far will it travel horizontally before landing again on the field? A. 45 m B. 60 m C. 79 m D. 150 m
C; To answer horizontal range questions, start by determining the time of flight. For that, focus on the vertical component of motion, picking a location where there is enough information to solve for the time. This occurs at the apex: . Given the time of flight, the range is a matter of distance = rate × time, where the rate is the horizontal, cosine component of velocity and total time is twice tapex: R = 30 m/s (cos30°) × 3 s ≈ 90 m (.85) ≈ 77 m, closest to choice C.
Two bricks are released simultaneously from the same height above the surface of the Earth. Brick #1 is simply dropped, while Brick #2 is given a purely horizontal initial velocity of magnitude 10 m/s. If they both strike the ground in 3 seconds, how far from Brick #1 will Brick #2 land? (Ignore air resistance.) A. 45 m B. 60 m C. 30 m D. 15 m
C; Two bricks are released simultaneously from the same height above the surface of the Earth. Brick #1 is simply dropped, while Brick #2 is given a purely horizontal initial velocity of magnitude 10 m/s. Ignoring air resistance, if they both strike the ground in 3 seconds, Brick #1 will land 30 mfrom Brick #2. Brick #1 will have no horizontal displacement, so the difference between them will be Brick #2's horizontal displacement: delta X = voxt = (10)(3) = 30m
A speed of 1 mi/h is equivalent to x ft/min. What is x? (Use 1 mi = 5280 ft.) A. 0.011 B. 0.66 C. 60 D. 88
D
An object undergoing a nonzero acceleration can be: A. slowing down. B. speeding up. C. turning at constant speed. D. all of these.
D
The linear density of a certain homogeneous metal bar is found to be 105 mg/cm. Express this density in kg/m. A. 10^3 kg/m B. 10^-3 kg/m C. 10^-2 kg/m D. 10^1 kg/m
D
Through free space, light travels at a speed of 3 × 108 m/s. Express this speed in kilometers per microsecond. A. 3 × 10^3 km/µs B. 300 km/µs C. 3 × 105 km/µs D. 0.3 km/µs
D
Which of the following kinematic descriptions is physically impossible? A. An object in motion for some period of time has zero average velocity but positive average speed. B. An object maintains constant speed but has nonzero acceleration. C. An object in motion has zero displacement but positive total distance traveled. D. An object in motion for some period of time has an average velocity with a greater magnitude than its average speed.
D
An object is dropped from a height of 980 m. Let T be the time required to fall the entire distance and let t be the time required to fall the first half of the distance. Calculate the ratio T/t. (Ignore air resistance.) A. 2 B. 1/2 C. 4 D. sqrt (2)
D;
Calculate the magnitude of your displacement after walking 3 mi north, then 6 mi west, then 5 mi north. A. 6 mi B. 14 mi C. 8 mi D. 10 mi
D;
A man pushes down at an angle of 60° from the horizontal on an initially stationary 100 kg crate on a hockey rink floor (no friction) with a force of 200 N. If the man applies this force for 3 seconds, how far does the crate move? A. 450 m B. 9 m C. 6 m D. 4.5 m
D; F(push) = ma 200 x cos 60 = ma 100 = 100a a = 1 m/s^2 x = ?, v0 = 0, t = 3, a = 1 m/s^2 x = vot + 1/2 at^2 x = 1/2at^2 x = (.5)(1)(9) x = 4.5 m
An object is released from rest at height h above the surface of the Earth, where h is much smaller than the radius of the Earth. It takes t seconds to fall to the ground. At what height should this object be released from rest in order to take 2t seconds to fall to the ground? (Ignore air resistance; g = magnitude of gravitational acceleration.) A. 2gh B. 4gh C. 2h D. 4h
D; An object is released from rest at height h above the surface of the Earth, where h is much smaller than the radius of the Earth. It takes t seconds to fall to the ground. Ignoring air resistance and taking g = magnitude of gravitational acceleration, the object should be released from a height of 4h from rest in order to take 2t seconds to fall to the ground. This is a proportion question (the MCAT loves proportions). Essentially it's asking, "What should happen to the height (distance) if the time is doubled?" We need to find the connection between height (distance) and time. With "down" positive—so a = g is positive—and with v0 = 0, Big Five #3 tells us that delta Y = 1/2at^2 h is proportional to t^2 4h
A car drives to a location with an average speed of 30 m/s and makes the return trip with an average speed of 40 m/s. What is the magnitude of the car's average speed for the entire trip? A. 36 m/s B. 35 m/s C. 0 m/s D. 34 m/s
D; The average speed cannot simply be found by averaging the two portions of the trip - the answer is not 35 m/s. (Note that if the car were uniformly accelerating from 30 m/s to 40 m/s, the average speed would then be 35 m/s.) Average speed is defined as the total distance divided by the time. Since the total distance was not given, we could say that car traveled a distance D there and a distance D back. Since distance equals average speed multiplied by the time, the time for the first half of the journey is D / 30 and the time for the second half of the journey is D / 40. Therefore, the average speed would be 2D / ((D / 30) + (D / 40)). A simpler way to solve this would be to choose a value for D, say 120 m. In this case, the average speed would be 2(120) / ((120 / 30) + (120 / 40)) which equals 240 / 70, or roughly 34 m/s. Conceptually speaking, just like the average test score is "weighted" toward the score that occurs most often, average speed is weighted toward the portion of the journey that takes the longest. Since the 30 m/s segment takes the longest, it should count more than the 40 m/s segment. The average speed, therefore, should be less than 35 m/s. 34 m/s is the only non-zero answer choice that is less than 35 m/s.
Which of the following kinematic descriptions is physically impossible? A. An object in motion for some period of time has zero average velocity but positive average speed. B. An object maintains constant speed but has nonzero acceleration. C. An object in motion has zero displacement but positive total distance traveled. D. An object in motion for some period of time has an average velocity with a greater magnitude than its average speed.
D; This question can be answered using process of elimination. Choices A and C are both possible (and therefore eliminated as answer choices) because an object traveling in a closed curved path (like a circle) will have zero displacement and therefore zero average velocity but not zero distance or average speed. Choice B is possible, therefore eliminated, because it is the case for uniform circular motion. Hence, choice D must be correct. Average velocity has magnitude of displacement / time, whereas average speed equals total distance / time. The time is the same for both, and displacement ≤ total distance, so average velocity ≤ average speed. Put another way, there's no shorter path than a straight line, for which the magnitude of displacement equals the total distance. For any other type of path, the displacement is shorter than the distance traveled.
The slope at a point on a velocity versus time graph represents what physical quantity? A. Displacement B. Instantaneous velocity C. Instantaneous acceleration D. Average acceleration
Instantaneous acceleration
Is it possible for cars to have the same velocity but different speeds?
No
Velocity tells us what two things?
Speed AND direction
For determining a projectile's vertical motion, use ________ theta
sin
When kinematics is graphed, time is always on the ____ axis
X
To calculate flight time in a projectile motion equation, use the ____ direction to get the time.
Y
Can an object be accelerating even if its speed is constant?
Yes; you can be traveling at the same speed but change direction (for example, turn left or right)
When acceleration is perpendicular to the velocity, the object's speed is
constant
For determining a projectile's horizontal velocity, use __________ theta
cosine
Kinematic equation (no acceleration)
d = 1/2(v0 + v) t
Kinematics equation (no v)
d = vot + 1/2 at^2
Kinematics equation (no v0)
d = vt - 1/2 at^2
What does acceleration tell us?
how quickly velocity changes
Displacement of an object is its change in
position
Distance is a
scalar
Average velocity equation
v = ∆x / ∆t
Kinematics equation (no time)
v^2 = v0^2 + 2ad
Displacement is a
vector
cos 45
√2/2
sin 45
√2/2
cos 30
√3/2
sin 60
√3/2