A car and train move together along parallel paths at 25,0 m/s, with the car
adjacent to the rear of the train. Then, because of red light , the car undergoes
a uniform acceleration of -5 m/s2 and come a rest. It remains at rest for 45 s
and the acceleration back to a speed of 20 m/s at rate of 2.5 m/s2
. How far
behind the rear of the train is the car when it reaches the speed of 25 m/s,
assuming that the speed of the train has remained 25,0 m/s?
To find the distance behind the rear of the train when the car reaches a speed of 25 m/s, we need to calculate the time it takes for the car to reach that speed and then use this time to calculate the distance traveled.
Let's break down the problem into parts:
1) The car undergoes a uniform acceleration of -5 m/s^2 until it comes to rest.
Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken, we can rearrange the equation to solve for time:
0 = 25 + (-5)t
5t = 25
t = 5 s
Therefore, it takes 5 seconds for the car to come to rest.
2) The car remains at rest for 45 seconds.
3) The car then accelerates back to a speed of 20 m/s at a rate of 2.5 m/s^2.
Again, using the equation of motion: v = u + at, and rearranging for time:
20 = 0 + 2.5t
2.5t = 20
t = 8 s
Therefore, it takes 8 seconds for the car to reach a speed of 20 m/s from rest.
4) Now, we need to find the time it takes for the car to accelerate from 20 m/s to 25 m/s.
Using the same equation of motion: v = u + at, where v is the final velocity (25 m/s), u is the initial velocity (20 m/s), and a is the acceleration (2.5 m/s^2), we can rearrange the equation to solve for time:
25 = 20 + 2.5t
2.5t = 5
t = 2 s
Therefore, it takes 2 seconds for the car to accelerate from 20 m/s to 25 m/s.
5) Finally, we can calculate the distance traveled during this time.
The distance traveled can be calculated using the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Since the initial velocity, u, is 20 m/s and the time, t, is 2 seconds, we can calculate the distance traveled:
s = (20)(2) + (1/2)(2.5)(2)^2
s = 40 + (1/2)(2.5)(4)
s = 40 + 5
s = 45 meters
Therefore, when the car reaches a speed of 25 m/s, it will be 45 meters behind the rear of the train, assuming the speed of the train remained at 25 m/s.