PHY 151 Homework Solutions 02A
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Average Velocity, and Speed
1.
A racing car starts from rest at t = 0 and reaches a final speed v at time t. If the
acceleration of the car is constant during this time, which of the following
statements are true?
(a)
The car travels a distance = vt
(b)
The average speed of the car is v/2
(c)
The magnitude of the acceleration of the car is v/t
(d)
The velocity of the car remains constant
(e)
None of statement (a) through (d) is true
2.
A person walks first at a constant speed of 5.00 m/s along a straight line from
point A to point B and then back along the line from B to A at a constant speed of
3.00 m/s.
(a)
What is her average speed over the entire trip?
(b)
What is her average velocity over the entire trip?
(a)
Average speed = total distance / total time
Let t be the time from A to B and t’ the time from B to A.
The distance from A to B is the same as the distance from B to A.
The distance is d = vt.
Therefore 5.00t = 3.00 t’
Solve for t’ = (5/3)t
Average speed = (5t + 3t’)/(t + t’)
Substitute for t’ gives
(5t + 5t)/(t + 5/3t) = 10/(8/3) = 30/8 = 3.75 m/s
(b)
The starting and ending points are at the same place. The displacement is zero. Therefore,
average velocity is 0 for the entire trip.
Instantaneous Velocity and Speed
3.
A position-time graph for a particle moving along the x axis is shown below.
(a)
Find the average velocity in the time interval t= 1.50 s to t = 4.00 s.
(b)
Determine the instantaneous velocity at t = 2.00 s.
(c)
At what value of t is the velocity zero?
(a) At t=1.50 s x = 8 m and at t = 4.00 s x = 2m
Average velocity is (2-8)/(4 – 1.5) = -6/2.5 = -2.4 m/s
(b) Instantaneous velocity at t = 2 is the slope at t = 2. The upper corner of green line is x
= 13 m and t = 0 s. The lower corner of green line is x = 0 m and t = 3.5 s
Slope = (13 – 0) / (0 – 3.5) = -3.7 m/s
(c)
The velocity is zero at t = 4 s.
4.
The position of a particle moving along the x axis varies in time according to the
expression x = 3t2, where x is in meters and t is in seconds. Evaluate its position
(a)
at t = 3.00 s
(b)
at 3.00 s + t
(c)
Evaluate the limit x/t as t approaches zero to find the velocity at t =
3.00 s.
Introductory Calculus Problem
(a) At t =3.00 s, x = 3(3)2 = 27 m
(b) At 3.00 s + t, x + x = 3(3 + t)2 = 3(9 + 6t + t2) = 27 + 18t + 3(t)2
(c) x / t = 18t + 3(t)2 / t = 18 + 3t
As t approaches zero and t approaches 3, x / t approaches 18
Particle under Constant Velocity
5.
A hare and a tortoise compete in a race over a straight course 1.00 km long. The
tortoise crawls at a speed of 0.200 m/s toward the finish line. The hare runs at
speed of 8.00 m/s toward the finish line for 0.800 km and then stops to tease the
slow-moving tortoise as the tortoise eventually passes by. The hare waits for a
while after the tortoise passes and then runs toward the finish line again at 8.00
m/s. Both the hare and the tortoise cross the finish line at the exact same instant.
Assume both animals, when moving, move steadily at their respective speeds.
(a)
How far is the tortoise from the finish line when the hare resumes that
race?
(b)
For how long in time was the hare stationary?
Not the shortest solution.
(a)
Time for tortoise to reach finish line.
d = vt
t = d/v = 1.00 km / 0.200 m/s = 1000 m / 0.200 m/s = 5000 s.
Time for hare to reach finish line.
First part of trip.
t = d/v = 0.800 km / 8 m/s = 800 m / 8 m/s = 100 s
Then hare waits for a time T.
Second part of trip.
t = d/v = 200 m / 8 m/s = 25 s
The total times of hare and tortoise are the same so
5000 s = 100 s + T + 25 s = 125 s + T
T = 5000 – 125 s = 4875 s
The hare resumes the race at 100 + 4875 s = 4975 s.
Distance of tortoise from finish line is 1000 – vt = 1000 – (0.200 x 4975) = 5 m.
(b)
Hare wait time is 4875 s.
Acceleration
6.
A 0.0500-kg Super Ball traveling at 25.0 m/s bounces off a brick wall and
rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in
contact with the wall for 0.00350 s, what is the magnitude of the average
acceleration of the ball during this time interval?
Average acceleration = (vf – vi)/t
Let the initial motion be in the +x direction.
Average acceleration = (-22 – 25)/0.00350 = -47/0.00350 = -1.34 x 104 m/s2
7.
When the pilot reverses the propeller in a boat moving north, the boat moves with
acceleration directed south. Assume the acceleration of the boat remains constant
in magnitude and direction. What happens to the boat:
(a)
It eventually stops and remains stopped.
(b)
It eventually stops and then speeds up in the forward direction.
(c)
It eventually stops and then speeds up in the reverse direction.
(d)
It never stops but loses speed more and more slowly forever.
(e)
It never stops but continues to speed up in the forward direction.
8.
An object moves along the x axis according to the equation x = 3.00t2 – 2.00t +
3.00, where x is in meters and t is in seconds. Determine
(a)
the average speed between t = 2.00s and t = 3.00s.
(b)
the instantaneous speed at t = 2.00 s and t = 3.00s
(c)
the average acceleration between t = 2.00 s and t = 3.00 s
(d)
the instantaneous acceleration at t = 2.00 s and t = 3.00 s
(e)
At what time is the object at rest?
Introductory calculus problem
(a)
t = 2.00, x = 3(2)2 – 2(2) + 3 = 12 – 4 + 3 = 11.0 m
t = 3.00, x = 3(3)2 – 2(3) + 3 = 27 - 6 + 3 = 24.0 m
average speed = (xf – xi) / t = (24 – 11)/1 = 13.0 m/s
(b)
x = 3.00t2 – 2.00t + 3.00
v = dx/dt = 6.00t – 2.00
t = 2.00, v = 6(2) – 2 = 10.0 m/s
t = 3.00, v = 6(3) – 2 = 16.0 m/s
(c)
average acceleration = (vf – vi)/t = (16-10)/1 = 6.00 m/s2
(d)
v = dx/dt = 6.00t – 2.00
a = dv/dt = 6.00 m/s2 at both times
(e)
v = dx/dt = 6.00t – 2.00 = 0
t = 2.00/6.00 = 0.333 s velocity is zero.