PHY202 Midterm Practice

-Drawing this problem shows that the bird's total displacement is equivalent to the resulting vector so... a^2 + b^2 = c^2 4^2 + 3^2 = c^2 c^2 = 25 c = 5 m

A bird flies 4 m south and 3 m west. What is the magnitude of the bird's total displacement?

What do we know? Δx = WANT TO FIND v0 = 5.0 m/s (to the right makes it positive) v = DO NOT HAVE a = -4.0 m/s/s (leftward makes it negative) t = 6.0 s -the only kinematics equation that includes Δx and excludes v is: Δx = (v0)t + (1/2)a(t^2) Δx = (5.0 m/s)(6.0 s) + (1/2)(-4.0 m/s/s)((6.0 s)^2) Δx = 30.0 m + (-72.0 m) Δx = -42 m

A boat is drifting to the right with a speed of 5.0 m/s when the driver turns on the motor. The motor runs for 6.0 s causing a constant leftward acceleration of magnitude 4.0 m/s/s. What is the displacement of the boat over the 6.0 s time interval? Use two sig figs.

-the velocity would begin positive and then have a steady negative slope until it becomes negative

A cannonball is fired upward and to the right from the edge of a cliff, at an angle of 20 degrees with the horizontal at time t = 0. We can ignore air resistance. What would the graph of the cannonball's vertical velocity look like over time?

What do we know? *everything is rightward so everything is positive Δx = 3.0 m v0 = 2.0 m/s v = 0 m/s (a complete stop) a = NOT GIVEN t = WANT TO FIND -the only kinematics equation the includes t and excludes a is: Δx = ((v + v0)/2)t 3.0 m = ((0 m/s + 2.0 m/s)/2)t 3.0 m = (1.0 m/s)t t = 3.0 s

A car is traveling to the right with a speed of 2.0 m/s on an icy road when the brakes are applied. The car slows down with a constant acceleration for 3.0 m until it comes to a stop. How long does it take the car to slide to a stop? Use two sig figs.

What do we know? *everything is rightward so everything is positive Δx = 110 m v0 = 29 m/s v = 34 m/s a = WANT TO FIND t = NOT GIVEN -the only kinematics equation that excludes t while including a is: v^2 = (v0)^2 + 2aΔx (34 m/s)^2 = (29 m/s)^2 + 2a(110 m) 1156 (m/s)^2 = (841 (m/s)^2) + a(220 m) 315 (m/s)^2 = a(220 m) a = 1.4 m/s/s

A car is traveling to the right with a speed of 29 m/s when the driver slams on the accelerator to pass a truck. With constant acceleration, the car passes the truck in 110 m and reaches a speed of 34 m/s. What was the acceleration of the car as it sped up? Use two sig figs.

(1)________________________________________ a = Δv/Δt = (0 m/s - 8.0 m/s)/(24 s - 12 s) = (-8.0 m/s)/(12 s) = -0.67 m/s/s (2)________________________________________ a = Δv/Δt = (8 m/s - 0 m/s)/(12 s - 0 s) = (8 m/s)/(12 s) = 0.67 m/s/s

A cat chases a mouse until the cat gets tired. (1) What is the cat's average acceleration from 12 to 24 seconds? (2) From 0 to 12 seconds?

-since the acceleration is constant at 4 m/s/s, the resulting velocity time graph would have a graph with a single straight line with a constantly increasing positive slope

A cheetah chases a gazelle, reaching a speed of 28 m/s. Describe the resulting velocity time graph.

_Draw this problem and see that 60 degrees is complimentary to the triangle's 30 degree theta, with the hypotenuse at 50 m -Since we want the horizontal displacement, we want the length of the triangle opposite of theta SOHCAHTOA sin(30) = x/50 m 50 m(sin(30)) = x x = 25 m

A chimp climbs a vine up and to the right for a total displacement of 50 m that makes a 60 degree angle from the ground. What was the horizontal displacement of the chimp?

What do we know? *everything is rightward so everything is positive Δx = 3.0 m v0 = 1.5 m/s v = WANT TO FIND a = 12 m/s/s t = DO NOT HAVE -the only kinematics equation that includes v0 and excludes t is: v^2 = (v0)^2 + 2aΔx v^2 = (1.5 m/s)^2 + 2(12 m/s/s)(3.0 m) v^2 = (2.25 (m/s)^2) + (72 (m/s)^2) v^2 = 74.25 (m/s)^2 v = 8.6 m/s

A dog walking to the right with a speed of 1.5 m/s sees a cat and speeds up with constant rightward acceleration of magnitude 12 m/s/s. What is the velocity of the dog after speeding up for 3.0 m? Use two sig figs.

*it asks for displacement so take into account that the frog jumps to the left (-1.8 m/s)(0.55 s) 0.99 m

A frog jumps to the left with an average speed of 1.8 m/s for 0.55 s. What was the frog's displacement in meters? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

an upside-down parabola

A golfer hits a golf ball upwards at an angle of 75 degrees with the horizontal. We can ignore air resistance. What would the vertical displacement of the ball over time look like on a graph?

-from the graph, we can see that there is no slope at t= 6s, so the instantaneous speed is 0 m/s *The slope of a given point is the instantaneous velocity. -also from the graph we can see that x = 0 m at t = 0s and also x = 1.0 m and t = 2 s v = (1.0 m - 0 m)/(2.0 s - 0s) v = 0.50 m/s

A lost bird flies forward and backward in a straight line. What is the instantaneous speed of the bird at t = 6 s? What is the instantaneous velocity at t = 1 s?

(1)________________________________________ -the displacement, Δx, can be considered the area under the time interval in a velocity time graph --since from t = 2 s to t = 3 s there is a slope, we can create "vertical asymptote" at t = 3 s so that we may find the area of a triable on the left side and the area of a rectangle from t= 3 s to t = 4s. The resulting displacement will sum the two areas: Δx = A1 + A2 -lets make A1 the triangle side and A2 the rectangular side --we also know that the area of a triangle can be described as A = (1/2)(base)(height) ---for a velocity-time graph, the base is the time along the x-axis and the height is the velocity along the y-axis ---so we have: A1 = (1/2)(3.0 s - 2.0 s)(0 - 5 m/s) A1 = (1/2)(1.0 s)(-5 m/s) A1 = -2.5 m --the area of a rectangle/square can be described as A = lw ---so we have: A2 = (4.0 s - 3.0 s)(-5 m/s) A2 = 1 s(-5 m/s) A2 = -5 m -FINALLY: Δx = A1 + A2 Δx = -2.5 m - 5.0 m Δx = -7.5 m (2)________________________________________ Δx = 0.50bh = 0.50vt = 0.50(1.0 s)(-5.0 m/s) = -2.5 m (3)________________________________________ -The area of our velocity vs. time graph is displacement Δx. Returning to the initial position comes at the time where the positive and negative areas cancel out so Δx = 0. -from the graph, we can see that the velocity becomes constant at t = 3.0 s, so what we are looking for is how much time after t = 3.0s is needed for Δx to = 0? -let's first find the displacement Δx up to t = 3.0 s, where A1 is from t = 0 s to t = 2.0 s (the first triangle on the left) and A2 is from t = 2.0 s to t = 3.0 s Δx = A1 + A2 A1 = 0.50(2.0 s)(5.0 m/s) = 5 m/s A2 = 0.50(1.0 s)(-5.0 m/s) = -2.5 m/s Δx = 5 m/s + (-2.5 m/s) = 2.5 m/s -now we want to know how long after t = 3.0 s Δx will = -2.5 m (since Δx = +2.5 m, and we need it to = 0 m) Δx = A = lw = tv -2.5 m = t(-5.0 m/s) t = 0.5 s -since the question asks at what time we need to add this time after 3.0 s to 3.0s therefore Δx = 0 at 3.5 s

A lost chicken is trying to cross the street. (1) What is the chicken's displacement Δx from t = 2s to 4s? (2) What is the displacement Δx from t = 2s to 3s? (3) At what time does the chicken have the same position as t = 0s?

v(avg) = Δx/Δt = (+20 m + (-5 m) + (-25 m))/300 s v(avg) = -10 m/300 s = -0.033 m/s avg speed = distance/Δt = (|20 m| + |-5 m| + |25 m|)/300 s avg speed = 50 m/300 s = 0.17 m/s

A pig runs rightward 20m to eat a juicy apple. It then walks leftward 5m to eat a nut. Finally, it walks leftward another 25m to eat another nut. The total time taken by the big was 300 seconds. What was the pig's average velocity and average speed over this time? -round to two sig figs

-the displacement, Δx, can be considered the area under the time interval in a velocity time graph -since we are looking at t= 0s to t = 1.5 s, we can see there are two areas underneath the graph on either side of the origin since the function goes from negative to positive, the resulting displacement will sum the two areas: Δx = A1 + A2 -both areas under the graph form triangles --we also know that the area of a triangle can be described as A = (1/2)(base)(height) ---SO WE HAVE A1 = 0.50(1.0 s - 0 s)(0 m/s - 2.0 m/s) A1 = 0.50(1.0 s)(-2.0 m/s) A1 = -1.0 m A2 = 0.50(1.5 s - 1.0 s)(1.0 m/s - 0 m/s) A2 = 0.50(0.50 s)(1.0 m/s) A2 = 0.25 m Δx = -1.0 m + 0.25 m Δx = -0.75 m

A robin is flying around. What is it's displacement Δx from t = 0s to 1.5s? Use two sig figs.

CHOOSING KINEMATICS EQUATIONS #1: v = v0 + at #2: Δx = (v0)t + (1/2)a(t^2) #3: Δx = ((v + v0)/2)t #4: v^2 = (v0)^2 + 2aΔx FIRST ask yourself, what are Δx, v0, v, a, and t What do we know? Δx = yes (41 m) v0 = NOT GIVEN, ASKED TO FIND v = yes (35 m/s) a = NOT GIVEN t = yes (2.3 s) -since we do not have a and it does not ask for it, we can rule out equations 1, 2, and 4, leaving #3 Δx = ((v + v0)/2)t

A roller coaster speeds up with constant acceleration for 2.3 s until it reaches a velocity of 35 m/s over 41 m. We want to find the initial velocity of the roller coaster before it started to accelerate. What kinematic formula would be most useful to solve for the target unknown?

*The slope of a given point is the instantaneous velocity. -from the graph, I can see that x = 3 m at t = 0 s and also x = 2.5 m at t = 2 s v = (2.5 m - 3.0 m)/(2.0 s - 0 s) v = -0.25 m/s *since it asks for SPEED we do not want the negative sign instantaneous speed = 0.25 m/s

A rooster struts in a straight line on a roof. What is the instantaneous speed of the rooster at t = 1 s?

-from the problem, we know... v0 = 0 m/s a = a(freeffall) = -g = -9.8 m/s/s t = 1.8 s -we do not know the displacement (in this case, Δy), and we want to find v, so we can use: v = v0 +at = 0 m/s + (-9.8 m/s/s)(1.8 s) = -18 m/s

A sleepy student drops a calculator out of a window. We can ignore air resistance. What is the velocity of the calculator after falling for 1.8 s? Use two sig figs.

v(avg) = (5 m - 2 m)/(10 s - 4 s) 3 m/6 s 0.5 m/s *The slope of a given point is the instantaneous velocity. -from the graph, we can see that there is no slope at t= 6s, so the instantaneous velocity is 0 m/s

A spider crawls vertically on a wall and its motion is show in the figure. What is the average velocity between t = 4s and t = 10s? What is the instantaneous velocity at time t = 6s?

we know... Δy = -4.0 m a = a(freefall) = -g = -9.8 m/s/s v0 = 0 m/s let's use: Δy = (v0)t + (1/2)a(t^2) -4.0 m = (0 m/s)t + (1/2)(-9.8 m/s/s)(t^2) -4.0 m = (-4.9 m/s/s)(t^2) t^2 = 0.816326531.... s^2 t = 0.90 s

A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0 m before it lands on the dog. We can ignore air resistance. How many seconds did the acorn fall? Use two sig figs.

v(avg) = (3 m - 10 m - 3 m)/12 s = -10 m/12 s = -0.83 m/s avg speed = (3 m + 10 m + 3 m)/12 s = 16 m/12 s = 1.3 m/s

A squirrel is looking for nuts. It first runs rightward 3 m and runs leftwards another 10 m to a second nut, and then runs 3 m leftward to a third nut. the total time spent running was 12 s. Find both average velocity and average speed. Use two sig figs.

CHOOSING KINEMATICS EQUATIONS #1: v = v0 + at #2: Δx = (v0)t + (1/2)a(t^2) #3: Δx = ((v + v0)/2)t #4: v^2 = (v0)^2 + 2aΔx FIRST ask yourself, what are Δx, v0, v, a, and t What do we know? Δx = NOT GIVEN, ASKED TO FIND v0 = given v = given (complete stop) a = given t = NOT GIVEN -not having t rules out equations 1, 2, and 3, leaving #4 v^2 = (v0)^2 + 2aΔx

A train is traveling at a speed of 80 km/h when the conductor applies the brakes. The train slows with a constant acceleration of magnitude 0.5 m/s/s. We want to find the distance the train travels from the time the brakes are applied until the train comes to a complete stop. Which kinematic formula would be most useful to solve for the target unknown?

(1) the second half of an upside-down parabola (2) the acceleration would be 0 (horizontal line at origin)

An adventurous cliff jumper runs horizontally off a cliff at time t = 0. We can ignore air resistance. (1) What would the graph of the cliff jumper's vertical displacement look like over time? (2) What would the graph of the cliff jumper's horizontal acceleration look like over time? Assume cliff as origin.

Draw this problem and see that 1 meter is the opposite of theta and 2 meters is the adjacent of theta... SOHCAHTOA tan(theta) = 1 m/2 m tan inv(1/2) = theta theta = 26.6 degrees

An ant walks 2 m to the right and then climbs 1 m up a wall. How many degrees above the ground is the ant compared to its initial position?

we know... v0 = 0 m/s a = a(freefall) = -g = -9.8 m/s/s t = 1.5 s -we do not know v, and we want to find the displacement (in this case, Δy), so we can use: Δy = (v0)t + (1/2)a(t^2) Δy = (0 m/s)(1.5 s) + (1/2)(-9.8 m/s/s)((1.5 s)^2) Δy = (-4.9 m/s/s)(2.25 s^2) Δy = -11 m

An egg is dropped from a treehouse and takes 1.5 seconds to reach the ground. We can ignore air resistance. What is the egg's displacement? Use two sig figs.

-from 0 to 10 s the velocity is constant, so there is no acceleration

Bill drives and sees a red light and slows to a stop. What is his average acceleration from 0 to 10 seconds?

CHOOSING KINEMATICS EQUATIONS #1: v = v0 + at #2: Δx = (v0)t + (1/2)a(t^2) #3: Δx = ((v + v0)/2)t #4: v^2 = (v0)^2 + 2aΔx FIRST ask yourself, what are Δx, v0, v, a, and t (1)________________________________________ Δx = we do not know v0 = rest, 0 m/s v = 80.0 km/h a = 1.45 m/s/s t = ASKS US TO FIND -we don't know Δx and it does not ask for it, so we can rule out equations 2, 3, and 4, leaving #1 v = v0 + at t = (v - v0)/a (2)________________________________________ Δx = ASKS US TO FIND v0 = rest, 0 m/s v = ASKS US TO FIND a = 2.40 m/s/s t = 12.0 s -for the first part, we need Δx, and we know that we do not have the final velocity either, so we can rule out all equations but #2 Δx = (v0)t + (1/2)a(t^2) -after we find Δx, we can plug it back into really any of the equations, but if we didn't want to find Δx first, we could use equation #1 v = v0 + at

CHOOSING KINEMATICS EQUATIONS (1) A light-rail commuter train accelerates at a rate of 1.35 m/s/s. How long does it take to reach its top speed of 80.0 km/h, starting from rest? (2) While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s/s for 12.0 s. How far does the car travel during those 12.0 s? What is the car's final velocity?

we know... a = a(freefall) = -g = -9.8 m/s/s t = 0.75 s and since he catches the ball at the same height.... Δy = 0 m -we do not know v and we want to find v0, so we can use: Δy = (v0)t + (1/2)a(t^2) 0 m = (v0)(0.75 s) + (1/2)(-9.8 m/s/s)((0.75 s)^2) 0 m = (v0)(0.75 s) + (-4.9 m/s/s)(0.5625 s^2) v0(0.75 s) = 0 m - (-4.9 m/s/s)(0.5625 s^2) v0(0.75 s) = 0m - (-2.75625 m) v0 = 3.7 m/s

Chef Andy tosses an orange in the air, then catches it again at the same height. The orange is in the air for 0.75 s. We can ignore air resistance. What was the orange's velocity at the moment it was tossed into the air?

scalar- speed and distance vector- velocity and displacement *vectors include magnitude(size) AND direction

Classify the following as vector or scalar quantities: speed displacement distance velocity

*it asks for average speed so only use magnitudes 2.1 m/0.52 s 4.0 m/s

Find the avg speed in m/s if a lizard leaps 2.1 m to the left in 0.52 s. Use two sig figs.

3.0 m/1 sec = 1 sec/3.0 m (1 sec/3.0 m)(50 m) 17 s

Grant sprints 50 m to the right with an average velocity of 3.0 m/s. How many seconds did he sprint? Use two sig figs.

displacement

change in position

v(avg) = (30 m - 30 m)/75 s = 0 m/75 s = 0 m/s avg speed = (30 m + 30 m)/75 s = 60 m/75 s = 0.80 m/s

Jenna rushes 30 m toward her truck from her house. She notices she forgot her jacket and returns back to the house. Her total time travel 75 s. Find average velocity and average speed. Use two sig figs.

*average velocity tells us that we want displacement over time rather than distance v(avg) = (20 m - 15 m + 20 m)/75 s v(avg) = 25 m/75 s v(avg) = 0.33 s *average speed tells us that we want to use distance rather than displacement avg speed = (20 m + 15 m + 25 m)/75 s avg speed = 55 m/75 s avg speed = 0.73 m/s

Justin walks from the house to his truck on a windy day. He walks 20 m toward the truck, 15 m back to retrieve his wind-blown hat, and another 20 m to reach the truck. He walked a total time of 75 s. What is Justin's average velocity over the 75 s period? What is Justin's average speed over the 75 s period? Use two sig figs.

Distance = |2| + |-3-1| = 6 units Displacement = xf - xi = -2 - 0 = -2 units

On a number line, an object moves two units to the right, then three units to the left, then another unit to the left. Find the distance traveled and the displacement.

Displacement: x5 - x0 = 6 m - (-2 m) = 8 m Distance traveled: in this case, it is the same as the displacement

Refer to photo

Displacement = xf - xi = 0 m - 6 m = -6 m Distance traveled = |15 - 6| + |15| = 24 m

Refer to the photo to find the total distance traveled and the displacement.

(5.5 m/s)(3.3 s) 18 m

Robin jogs with an average speed of 5.5 m/s for 3.3 s. What was her distance travelled in meters? Use two sig figs

-the velocity is constantly decreasing, so there will be a constant negative velocity of a = (-15 m/s)/5 s = -3 m/s/s -therefore there will be a horizontal line at the y-coordinate a = -3 m/s/s

Superman runs until reaching his flight takeoff speed of 15 m/s. Describe the resulting acceleration time graph.

Acceleration remains constant and speed increases

Violet throws a ball down from a bridge and it strikes the water below. We can ignore air resistance. What is true about the acceleration and speed of the ball on the way down?

distance

total length of path