Physics 1 Test

A firkin is an old British unit of volume equal to 9 gallons. How may cubic meters are there in 8.60 firkins?

8.60 firkins = 8.60 firkins (9 gal/1 firkin) * (3.786 L/1 gal) * (10^3 cm^3/1 L) * (1 m^3/10^6 cm^3) = 0.293 m^3

Calculate the x-component and the y-component (in m) of the vector with magnitude 28.0 m and direction 41.0°.

A vector r with magnitude 28.0 m and direction 41.0° has the following x-component. x = r cos(𝜃) = (28.0 m) cos(41.0°) = 21.1 m A vector r with magnitude 28.0 m and direction 41.0° has the following y-component. y = r sin(𝜃) = (28.0 m) sin(41.0°) = 18.4 m

A particle starts from rest and accelerates as shown in the figure below. (a) Determine the particle's speed at t = 10.0 s and at t = 20.0 s. (b) Determine the distance traveled in the first 20.0 s. (Enter your answer to one decimal place.)263 m

From the figure below, observe that the motion of this particle can be broken into three distinct time intervals, during each of which the particle has a constant acceleration. These intervals and the associated accelerations are 0≤t < 10.0 s, a = a1 = +2.00 m/s2 10≤t < 15.0 s, a = a2 = 0 and 15.0≤t < 20.0 s, a = a3 = −3.00 m/s2. (a) Applying vf = vi + a(Δt) to each of the three time intervals gives the following. for 0 ≤ t < 10.0 s, v10 = v0 + a1(Δt1) = 0 + (2.00 m/s2)(10.0 s) = 20.0 m/s for 10.0 s ≤ t < 15.0 s, v15 = v10 + a2(Δt2) = 20.0 m/s + 0 = 20.0 m/sfor 15.0 s ≤ t < 20.0 s, v20 = v15 + a3(Δt3) = 20.0 m/s + (−3.00 m/s)(5.00 s) = 5.00 m/s (b) Applying Δx = vi(Δt) + 1/2a(Δt)2 to each of the time intervals gives to each of the time intervals the following. for 0 ≤ t < 10.0 s, Δx1 = v0Δt1 + 1/2a1(Δt1)2 = 0 + 1/2(2.00 m/s2)(10.0 s)(10.0 s)2 = 1.00 ✕ 10^2 m for 10.0 s ≤ t < 15.0 s,Δx2 = v10Δt2 + 12a2(Δt2)2 = (20.0 m/s)(5.00 s) + 0 = 1.00 ✕ 10^2 m for 15.0 s ≤ t < 20.0 s, Δx3 = v15Δt3 + 1/2a3(Δt3)^2 = 0 + (20.0 m/s)(5.00 s)+ 12(−3.00 m/s)(5.00 s)2 = 62.5 m Thus, the total distance traveled in the first 20.0 s is Δxtotal = Δx1 + Δx2 + Δx3 = 100 m + 100 m + 62.5 m = 263 m.

Newton's law of universal gravitation can be expressed by the equation F = G Mmr2, where F is the gravitational force, M and m are masses, and r is a length. Force has the SI units kg · m/s2. What are the SI units of the proportionality constant G?

From the universal gravitation law, the constant G is G = Fr2Mm. Its units are then [G] = [F][r2]/[M][m] = (kg · m/s2)(m2)/kg · kg = m3/kg · s2.

A truck covers 40.0 m in 8.60 s while uniformly slowing down to a final velocity of 1.75 m/s. Find the truck's original speed.7.55 m/s Find its acceleration.-0.675 m/s2

N/a

The speed of a nerve impulse in the human body is about 100 m/s. If you accidentally stub you toe in the dark, estimate the time it takes the nerve impulse to travel to your brain. (Assume that you are approximately 2.00 m tall and that the nerve impulse travels at uniform speed.)

We assume that you are approximately 2.00 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then Δt = Δx/v = 2.00 m/100 m/s = 2.00 ✕ 10−2 s = 0.0200 s.

A girl delivering newspapers covers her route by traveling 6.00 blocks west, 6.00 blocks north, and then 8.00 blocks east. (a) What is her final position relative to her starting location? (Enter the magnitude in blocks and the direction in degrees north of east.) magnitude6.32 blocksdirection71.6 ° north of east (b) What is the length (in blocks) of the path she walked? 20blocks The distance between two points depends only on the endpoints and not on the path connecting them.

(a) Set the origin at the girl's starting location with positive x-directed east and positive y-directed north. The girl's final x-position is the net effect of her east-west movements. x = −6.00 blocks + 8.00 blocks = 2.00 blocks Her final y-position is the following. y = 6.00 blocks The distance, d, from the origin to the girl's final position is therefore the following. d = (sqrt)(x)^2 + (y)^2 = (sqrt)(2.00 blocks)^2 + (6.00 blocks)^2 = 6.32 blocks The direction, 𝜃, from the origin to the girl's final position is the following. 𝜃 = tan^−1(y/x) = tan^−1(6.00 blocks/2.00 blocks) = 71.6° (b) The length, ℓ, of a path walked by the girl is the following. ℓ = 6.00 blocks + 6.00 blocks + 8.00 blocks = 20.0 blocks

The graph below plots velocity vs. time for an object moving along a straight path. (a) What is the average acceleration of the object during the following time intervals? (Enter your answers in m/s2. Indicate the direction with the signs of your answers.) (i) 0 s to 7.00 s= 0 m/s2 (ii) 7.00 s to 21.0 s= 3.43 m/s2 (iii) 0 s to 28.0 s= 1.71 m/s2 (b) What is its instantaneous acceleration at each of the following times? (Enter your answers in m/s2. Indicate the direction with the signs of your answers.) (i) 1.00 s= 0 m/s2 (ii) 14.0 s= 3.43 m/s2 (iii) 26.0 s= 0 m/s2

(a) The average acceleration over any time interval is a = Δv/Δt = vf − vi/(tf − ti) We will read the initial and final velocities and times directly from the graph. (i) From 0 to 7.00 s, a = (−24.0 m/s − (−24.0 m/s))/(7.00 s − 0) = 0. (ii) From 7.00 s to 21.0 s, a = (24.0 m/s − (−24.0 m/s))/(21.0 s − 7.00 s) = 3.43 m/s2. (iii) From 0 to 28.0 s, a = (24.0 m/s − (−24.0 m/s))/(28.0 s − 0) = 1.71 m/s2. (b) At any instant, the instantaneous acceleration equals the slope of the line tangent to the v vs. t graph at that point in time. (i) At t = 1.00 s, the slope of the tangent line is the same as the slope of the segment of the graph from 0 to 7.00 s, which is 0. (ii) At t = 14.0 s, the slope of the tangent line is the same as the slope of the segment of the graph from 7.00 s to 21.0 s. As we found above, the slope is 3.43 m/s2. (iii) At t = 26.0 s, the slope of the tangent line is the same as the slope of the segment of the graph from 21.0 s to 28.0 s, which is 0.

The graph below plots the position versus time for a particular body moving along the x-axis. What is the average velocity over the following time intervals (in m/s)? (Indicate the direction with the signs of your answers.)

(a) The average velocity over any time interval is v = Δx/Δt = (xf − xi)/(tf − ti). We will read the initial and final positions and times directly from the graph. From 0 to 2.00 s, v = (16.0 m − 0)/(2.00 s − 0) = 8.00 m/s. (b) From 0 to 4.00 s, v = (4.00 m − 0)/(4.00 s − 0) = 1.00 m/s. (c) From 2.00 s to 4.00 s, v = (4.00 m − 16.0 m)/(4.00 s − 2.00s) = −6.00 m/s. (d) From 4.00 s to 7.00 s, v = (−6.00 m − 4.00 m)/(7.00 s − 4.00s) = −3.33 m/s. (e) From 0 to 8.00 s, v = 0/(8.00 s − 0) = 0.

A figure skater glides along a circular path of radius 6.70 m. If she coasts around one half of the circle, find the following. (a) her distance (in m) from the starting location 13.4m (b) the length (in m) of the path she skated 21m The distance between two points depends only on the endpoints and not on the path connecting them.

(a) The endpoints of a path half way around a circle lie on a diameter. For a circle of radius R = 6.70 m the distance, d, between the points is therefore the following. d = 2R = 2(6.70 m) = 13.4 m (b) The length, ℓ, of a path half way around a circle equals half the circle's circumference. ℓ = 1/2(2𝜋R) = 𝜋R = 𝜋(6.70 m) = 21.0 m

A football player runs from his own goal line to the opposing team's goal line, returning to his ten-yard line, all in 28.5 s. Calculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.) (a) Calculate his average speed. 6.67 (b) Calculate the magnitude of his average velocity. 0.351 Average velocity is not the same as average speed. Average speed is calculated using the total path length traveled. yards/s

(a) The player runs 100 yards from his own goal line to the opposing team's goal line. Then he runs an additional 90 yards back to his ten-yard line, all in 28.5 s. Substitute values into the definition of average speed to find the following. Average speed = path length /elapsed time = (100 yards + 90 yards)/28.5 s = 6.67 yards/s (b) After returning to his ten-yard line, the player's displacement is 10 yards. Substitute values into the definition of average velocity to find the following. v = Δx/Δt = 10 yards/28.5 s = 0.351 yards/s

A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 60.0 min at 75.0 km/h, 5.0 min at 85.0 km/h, and 45.0 min at 55.0 km/h and spends 40.0 min eating lunch and buying gas. (a) What is the total distance traveled over the entire trip (in km)? 123.33 km (b) What is the average speed for the entire trip (in km/h)? 49.33 km/h

(a) We can first calculate the distances traveled. Note the speeds are in km/h, so we convert the times in minutes to hours. Δx1 = v1(Δt1) = (75.0 km/h)(60.0 min)(1 h/60 min) = 75.0 kmΔx2 = v2(Δt2) = (85.0 km/h)(5.0 min)(1 h/60 min) = 7.08 kmΔx3 = v3(Δt3) = (55.0 km/h)(45.0 min)(1 h/60 min) = 41.3 km The total distance traveled is then Δx = Δx1 + Δx2 + Δx3Δx = 75.0 km + 7.08 km + 41.3 km = 123 km. (b) The average speed for the trip is the total distance traveled divided by the total time. Note the total time includes the time spent at rest. Vav = Δx/Δt = Δx/Δt1 + Δt2 + Δt3 + Δt4 Vav = 123 km/(60.0 min + 5.0 min + 45.0 min + 40.0 min)(1 h/60 min) = 49.3 km/h

A motorist on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 85.0 km/h, 20.0 min at 100.0 km/h, and 60.0 min at 40.0 km/h and spends 25.0 min eating lunch and buying gas. (a) What is the total distance traveled over the entire trip (in km) =151km (b) What is the average speed for the entire trip (in km/h)? =56.7km/h

(a) We can first calculate the distances traveled. Note the speeds are in km/h, so we convert the times in minutes to hours. Δx1 = v1(Δt1) = (85.0 km/h)(55.0 min)(1 h/60 min) = 77.9 kmΔx2 = v2(Δt2) = (100.0 km/h)(20.0 min)(1 h/60 min) = 33.3 kmΔx3 = v3(Δt3) = (40.0 km/h)(60.0 min)(1 h/60 min) = 40.0 km The total distance traveled is then Δx = Δx1 + Δx2 + Δx3Δx = 77.9 km + 33.3 km + 40.0 km = 151 km. (b) The average speed for the trip is the total distance traveled divided by the total time. Note the total time includes the time spent at rest. vav = (Δx/Δt) = Δx/(Δt1 + Δt2 + Δt3 + Δt4) Vav = 151 km/(55.0 min + 20.0 min + 60.0 min + 25.0 min)(1 h/60 min) = 56.7 km/h

Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 65 mi/h. (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 5 mi away?0.839 min (b) How far must the faster car travel before it has a 15-min lead on the slower car?89.4

(a) The time for a car to make the trip is given by the following formula. t = Δx/v Thus, the difference in the times for the two cars to complete the same 5 mile trip is calculated as follows. Δt = t1 − t2 = Δx/v1 − Δx/v2 = (5 mi/55 mi/h) − (5 mi/65 mi/h)(60 min/1 h) = 0.839 min (b) When the faster car has a 15.0 min lead, it is ahead by a distance equal to that traveled by the slower car in a time of 15.0 min. This distance is given by the following equation. Δx1 = v2(Δt) = (55 mi/h)(15 min) The faster car pulls ahead of the slower car at the following rate. Vrelative = 65 mi/h − 55 mi/h = 10 mi/h Thus, the time required for it to get distance Δx1 ahead is the following. Δt = Δx/vrelative = (55 mi/h * 15min)/10 mi/h = 82.500 min Finally, the distance the faster car has traveled during this time is given below. Δx = vt = (65 mi/h)(82.500 min)(1 h/60 min) = 89.4 mi

A vector has an x-component of −28.5 units and a y- component of 42.5 units. Find the magnitude and direction of the vector.

Ax = −28.5 Ay = 42.5 A = (sqrt)Ax^2 + Ay^2 = (−28.5)^2 + (42.5)^2 = 51.2 units From the triangle, we find that 𝜑 = tan^−1(Ay/|Ax|) = tan^−1(42.5/28.5) = 56.2°, so 𝜃 = 180° − 𝜑 = 124°. Thus, A = 51.2 units at 124°counterclockwise from the +x-axis.

The map below shows that Atlanta is 730 miles in a direction 5.00° north of east from Dallas, and that Chicago is 560 miles in a direction 21.0° west of north from Atlanta. Assuming the Earth may be approximated as flat across this map, what is the displacement from Dallas to Chicago? Find its magnitude (in mi) and direction (in degrees) as measured counterclockwise from due east of Dallas.

Choose the positive x-direction to be eastward and the positive y-direction to be northward. The displacement vector R from Dallas to Chicago is shown in the figure below. Dallas, Atlanta, and Chicago are plotted on an x y coordinate plane where x is due east and y is due north. Dallas is at the origin. A vector labeled "730 mi" points rightwards from Dallas to Atlanta at an angle of 5° counterclockwise from due east. A vector labeled "560 mi" points upwards from Atlanta to Chicago at an angle of 21° counterclockwise from due north. A vector R points from Dallas to Chicago at an acute angle 𝜃 above the x-axis. Let vector A be the displacement from Dallas to Atlanta. The components of A are given by Ax = (730 mi)cos(5.00°) = 727 mi Ay = (730 mi)sin(5.00°) = 63.6 mi. Let vector B be the displacement from Atlanta to Chicago. Note the given angle is measured from the +y-direction, and note from the diagram that we can tell the x-component must be negative. Bx = −(560 mi)sin(21.0°) = −201 mi By = (560 mi)cos(21.0°) = 523 mi Since R = A + B, the components of R are given by Rx = Ax + Bx = 727 mi − 201 mi = 527 mi Ry = Ay + By = 63.6 mi + 523 mi = 586 mi. The magnitude of R is then R = (sqrt) Rx^2 + Ry^2 = (527 mi)^2 + (586 mi)^2 = 788 mi. The direction, measured counterclockwise from +x, is given by 𝜃 = tan^−1(Ry/Rx) = tan^−1(586 mi/527 mi) = 48.1° counterclockwise from due east of Dallas.

A certain car is capable of accelerating at a rate of 0.59 m/s2. How long does it take for this car to go from a speed of 25 mi/h to a speed of 30 mi/h?

From a = Δv/Δt, we have Δt = Δv/a = (30 − 25) mi/h/0.59 m/s2 x (0.447 m/s/1 mi/h) = 3.79 s

An object moving with uniform acceleration has a velocity of 13.0 cm/s in the positive x-direction when its x-coordinate is 2.85 cm. If its x-coordinate 3.55 s later is −5.00 cm, what is its acceleration? -8.57 cm/s2

N/a

The brakes are applied to a moving truck, causing it to uniformly slow down. While slowing, it moves a distance of 40.0 m in 7.00 s to a final velocity of 1.30 m/s, at which point the brakes are released. What was its initial speed (in m/s), just before the brakes were applied? 10.1 m/s What was its acceleration (in m/s2) while the brakes were applied? (Assume the initial direction of motion is the positive direction. Indicate the direction with the sign of your answer.) -1.26 m/s2

N/a

A house is advertised as having 1900 square feet under roof. What is the area of this house in square meters? HINT

Use the conversion factor 1 m = 3.281 ft to find the following. A = 1900 ft2 = (1900 ft2)(1 m/3.281 ft)2 = 176 m2

A tennis player moves in a straight-line path as shown in the figure below. Find her average velocity in the following time intervals. (a) 0 to 1.0 s=4 m/s(b) 0 to 4.0 s=-0.5 m/s(c) 1.0 s to 5.0 s=-1 m/s(d) 0 to 5.0 s= 0

The average velocity over any time interval is v = Δx/Δt = xf − xi/tf − ti) (a) v = Δx/Δt = (4.0 m − 0)/(1.0 s − 0) = +4.0 m/s (b) v = Δx/Δt = (−2.0 m − 0)/(4.0 s − 0) = −0.50 m/s (c) v = Δx/Δt = (0 − 4.0 m)/(5.0 s − 1.0 s) = −1.0 m/s (d) v = Δx/Δt = (0 − 0)/(5.0 s − 0 )= 0

Vector A has a magnitude of 36.5 units and it points in a direction 325° counterclockwise from the positive x-axis. What are the x- and y-components of A?

The x- and y-components of vector A are its projections on lines parallel to the x- and y-axes, respectively, as shown in the sketch. The magnitude of these components can be computed using the sine and cosine functions as shown below. Ax = |A| cos(325°) = +|A| cos(35.0°) = (36.5)cos(35.0°) = 29.9 units and Ay = |A| sin(325°) = −|A| sin(35.0°) = −(36.5)sin(35.0°) = −20.9 units.

The Roman cubitus is an ancient unit of measure equivalent to about 0.445 m. Convert the 2.09-m height of a basketball forward to cubiti.

Use the conversion factor 1 cubitus = 0.445 m to find the basketball player's height, h, in cubiti. h = 2.09 m = (2.09 m) *(1 cubitus /0.445 m) = 4.70 cubiti

Vector A has a magnitude of 33 units and points in the positive y-direction. Vector B is added to A, giving a resultant vector A + B that points in the negative y-direction with a magnitude of 16 units. What is the magnitude and direction of B?

We are given that R = A + B. When two vectors are added graphically, the second vector is positioned with its tail at the tip of the first vector. The resultant then runs from the tail of the first vector to the tip of the second vector. In this case, vector A will be positioned with its tail at the origin and its tip at the point (0, 33). The resultant is then drawn, starting at the origin (tail of first vector) and going 16 units in the negative y-direction to the point (0, −16). The second vector, B, must then start from the tip of A at point (0, 33) and end on the tip of R at point (0, −16) as shown in the sketch below. From this, it is seen that B is 49 units in the negative y-direction. magnitude 49 units and direction -y

A 50.0-g Super Ball traveling at 28.5 m/s bounces off a brick wall and rebounds at 15.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.40 ms, what is the magnitude of the average acceleration of the ball during this time interval? 12900 m/s2

We choose the positive direction to point away from the wall. Then, the initial velocity of the ball is vi = −28.5 m/s and the final velocity is vf = +15.5 m/s. If this change in velocity occurs over a time interval of Δt = 3.40 ms (i.e., the interval during which the ball is in contact with the wall), the average acceleration is a = Δv/Δt = vf − vi/Δt = +15.5 m/s − (−28.5 m/s)/3.40 ✕ 10−3 s = 1.29 ✕ 104 m/s2

A small turtle moves at a speed of 662 furlongs per fortnight. Find the speed of the turtle in centimeters per second. Note that 1 furlong = 220 yards and 1 fortnight = 14 days.

v = ℓt = (662 furlongs/1 fortnight) *(1 fortnight/14 days) * (1 day/8.64 ✕ 10^4s) * (220 yds/1 furlong) *(3 ft/1 yd) * (100 cm/3.281 ft) giving v = 11.0 cm/s.