An automobile and train move together along

parallel paths at 39.1 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s
2
because of a red lightand comes to rest. It remains at rest for 22.5 s,
then accelerates back to a speed of 39.1 m/s
at a rate of 2.53 m/s
2
.
How far behind the train is the automobile
when it reaches the speed of 39.1 m/s, assuming that the train speed has remained at
39.1 m/s?
Answer in units of m

times involved:

deacceleratin: 39.1/4 seconds
sittn still: 22 seconds
accelerating: 38.1/2.53 seconds

distance traveled: time deacc*39.1/2 + timeaccelerating*39.1/2

avg velocity= distance traveled/total time.

distance behind train: avgvelocitydifference*timetotalelapsed
=(39.1-avgvelocity)*totaltimeelpased.

To find the distance between the automobile and the train when the automobile reaches a speed of 39.1 m/s, we need to calculate the distance traveled by the train during the time the automobile was at rest.

Let's first find the time it takes for the automobile to come to rest under a uniform acceleration of -4 m/s^2.

Using the equation of motion:

v = u + at

where:
v = final velocity (0 m/s, as the automobile comes to rest)
u = initial velocity (39.1 m/s)
a = acceleration (-4 m/s^2)
t = time taken

Rearranging the equation to solve for t:

t = (v - u) / a

t = (0 - 39.1) / -4
t = 9.775 s

The automobile takes 9.775 seconds to come to rest.

Now, we need to find the distance traveled by the train during this time.

Using the formula:

s = ut + (1/2)at^2

where:
s = distance
u = initial velocity (39.1 m/s)
t = time (9.775 s)
a = acceleration (0, as the train maintains a constant speed)

s = (39.1)(9.775) + (1/2)(0)(9.775)^2
s = 381.5025 m

The train traveled a distance of 381.5025 meters while the automobile was at rest.

So, when the automobile reaches a speed of 39.1 m/s, it is 381.5025 meters behind the train.

To find the distance between the automobile and train when the automobile reaches a speed of 39.1 m/s, we need to calculate the distance traveled by both vehicles during the time the automobile is at rest.

First, let's calculate the time it takes for the automobile to accelerate from rest to a speed of 39.1 m/s. We can use the formula:

v = u + at

where:
v = final velocity (39.1 m/s)
u = initial velocity (0 m/s)
a = acceleration (2.53 m/s^2)
t = time

Rearranging the formula, we have:

t = (v - u) / a

t = (39.1 - 0) / 2.53
t = 15.43 s

Next, we calculate the distance traveled by the automobile during this time using the formula:

s = ut + (1/2)at^2

where:
s = distance
u = initial velocity (0 m/s)
a = acceleration (2.53 m/s^2)
t = time (15.43 s)

s = (0)(15.43) + (1/2)(2.53)(15.43)^2
s = 192.60 m

The automobile travels a distance of 192.60 m while accelerating from rest to a speed of 39.1 m/s.

Now, let's calculate the distance traveled by the train in the same time. Since the train maintains a constant speed of 39.1 m/s, the distance traveled can be calculated using the formula:

s = vt

where:
s = distance
v = velocity (39.1 m/s)
t = time (15.43 s)

s = (39.1)(15.43)
s = 603.47 m

The train travels a distance of 603.47 m during the time the automobile accelerates from rest to a speed of 39.1 m/s.

To find the distance between the two vehicles when the automobile reaches the same speed as the train, we subtract the distance traveled by the automobile from the distance traveled by the train:

Distance = Train distance - Automobile distance
Distance = 603.47 m - 192.60 m
Distance = 410.87 m

Therefore, the automobile will be 410.87 m behind the train when it reaches a speed of 39.1 m/s.