An automobile and train move together along
parallel paths at 21.8 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s
2
because of a red light
and comes to rest. It remains at rest for 19.1 s,
then accelerates back to a speed of 21.8 m/s
at a rate of 3 m/s
2
.
How far behind the train is the automobile
when it reaches the speed of 21.8 m/s, assuming that the train speed has remained at
21.8 m/s?
Answer in units of m
To find the distance between the automobile and the train when the automobile reaches a speed of 21.8 m/s, we need to consider the time it takes for the automobile to accelerate and reach that speed.
First, let's find the time it takes for the automobile to decelerate and come to a stop. We can use the equation:
v = u + at
Where:
v = final velocity (0 m/s as the automobile comes to a stop)
u = initial velocity (21.8 m/s)
a = acceleration (-4 m/s^2)
t = time taken to stop
Rearranging the equation to solve for t, we have:
t = (v - u) / a
t = (0 - 21.8) / -4
t = 21.8 / 4
t = 5.45 seconds
Next, we need to find the distance the automobile travels while at rest for 19.1 seconds. This is simply the product of the time and the velocity, as the automobile is not accelerating or decelerating during this time:
distance = velocity * time
distance = 0 m/s * 19.1 s
distance = 0 meters
Now, let's find the time it takes for the automobile to accelerate from rest back to a speed of 21.8 m/s. We can use the same equation as before, but with different values:
v = u + at
Where:
v = final velocity (21.8 m/s)
u = initial velocity (0 m/s, as the automobile is at rest)
a = acceleration (3 m/s^2)
t = time taken to accelerate
Rearranging the equation to solve for t, we have:
t = (v - u) / a
t = (21.8 - 0) / 3
t = 21.8 / 3
t = 7.27 seconds
Now, we can calculate the distance the automobile travels during this acceleration:
distance = velocity * time
distance = 21.8 m/s * 7.27 s
distance = 158.566 meters
Finally, to find the distance between the automobile and the train when the automobile reaches a speed of 21.8 m/s, we add up the distances traveled during deceleration, rest, and acceleration:
distance between automobile and train = 0 meters + 0 meters + 158.566 meters = 158.566 meters
Therefore, when the automobile reaches a speed of 21.8 m/s, it is approximately 158.566 meters behind the train.