A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 10.0 m/s and accelerates at the rate of 0.600 m/s2 for 7.00 s.
(a) What is his final velocity?
(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?
(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate and traveled at 10.2 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?
distance
time
a. V = Vo * a*t = 10 + 0.6*7 = 14.2 m/s.
b. T = d/r1 - d/r2 = 300/10-300/14.2 =
8.87 s. Saved.
To answer these questions, we can use the equations of motion. The first equation we will use is:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
(a) Final velocity can be calculated using the equation:
v = u + at
Given:
u = 10.0 m/s (initial velocity)
a = 0.600 m/s^2 (acceleration)
t = 7.00 s (time)
Substituting the given values into the equation, we have:
v = 10.0 m/s + (0.600 m/s^2)(7.00 s)
Simplifying, we get:
v = 10.0 m/s + 4.20 m/s
Therefore, the final velocity is:
v = 14.20 m/s
(b) To determine the time saved by accelerating at the end, we need to calculate the time it would have taken to cover the remaining distance without acceleration. We can use the equation:
s = ut + (1/2)at^2
Given:
u = 14.20 m/s (final velocity)
a = 0 m/s^2 (since the racer continued at this velocity)
s = 300 m (remaining distance)
Rearranging the equation, we get:
t = (2s) / (u + v)
Substituting the given values, we have:
t = (2 * 300 m) / (14.20 m/s + 14.20 m/s)
Simplifying, we get:
t = 600 m / 28.40 m/s
Therefore, the time saved is:
time saved = (time without acceleration) - (time with acceleration)
time saved = [(600 m) / (10.20 m/s)] - 7.00 s
(c) Now, let's calculate the distance the winner finished ahead of the other racer. We will use the equation:
s = ut + (1/2)at^2
Given:
u = 10.2 m/s (constant velocity of the other racer)
a = 0 m/s^2 (since the other racer did not accelerate)
t = 7.00 s (time of acceleration for the winner)
We need to calculate the distance traveled by the other racer during the acceleration time of the winner. Using the equation:
s = ut + (1/2)at^2
Substituting the given values, we get:
s = (10.2 m/s)(7.00 s) + (1/2)(0 m/s^2)(7.00 s)^2
Simplifying, we have:
s = 71.4 m
Therefore, the winner finished ahead of the other racer by:
distance = (remaining distance) + (distance traveled by the other racer during acceleration)
distance = 300 m + 71.4 m
For the time, we already know that both racers accelerated for 7.00 seconds, so they finish at the same time.
Therefore, the winner finished 371.4 meters ahead of the other racer.