An automobile and train move together along
parallel paths at 31.6 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s2 because of a red light
and comes to rest. It remains at rest for 44 s,
then accelerates back to a speed of 31.6 m/s
at a rate of 2.22 m/s2.
How far behind the train is the automobile
when it reaches the speed of 31.6 m/s, as-
suming that the train speed has remained at
31.6 m/s?
Answer in units of m
To find the distance behind the train, we need to determine the time it takes for the automobile to reach a speed of 31.6 m/s and then use that time to calculate the distance traveled.
First, let's find the time it takes for the automobile to accelerate from rest to 31.6 m/s.
Using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given:
Initial velocity (u) = 0 m/s (since the automobile is at rest)
Final velocity (v) = 31.6 m/s
Acceleration (a) = 2.22 m/s²
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (31.6 - 0) / 2.22
t = 31.6 / 2.22
t ≈ 14.23 seconds
Now that we have the time it takes for the automobile to reach a speed of 31.6 m/s, we can calculate the distance traveled during this time.
Using the equation: s = ut + 0.5at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Given:
Initial velocity (u) = 0 m/s (since the automobile was at rest)
Time (t) = 14.23 s
Acceleration (a) = 2.22 m/s²
Substituting the given values:
s = 0 × 14.23 + 0.5 × 2.22 × (14.23)²
s = 0 + 0.5 × 2.22 × (202.2529)
s ≈ 450.776 m
Therefore, the distance behind the train when the automobile reaches a speed of 31.6 m/s is approximately 450.776 meters.
To find the distance between the automobile and the train when the automobile reaches a speed of 31.6 m/s, we can break down the motion of the automobile into three segments:
1. The initial motion of the automobile at a constant velocity of 31.6 m/s while the train also moves at the same speed.
2. The deceleration of the automobile to come to rest at a rate of -4 m/s^2.
3. The acceleration of the automobile back to a speed of 31.6 m/s at a rate of 2.22 m/s^2.
Let's calculate the time and distance for each segment and then add them together to find the total distance.
1. Initial motion:
Since both the automobile and train move at a constant velocity of 31.6 m/s, they cover the same distance over any given time. Therefore, we do not need to consider this segment for finding the distance between the two.
2. Deceleration to come to rest:
The deceleration of the automobile is -4 m/s^2. To find the time it takes for the automobile to come to rest, we can use 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.
The final velocity is 0 m/s, the initial velocity is 31.6 m/s, and the acceleration is -4 m/s^2. Plugging these values into the equation, we can solve for t:
0 = 31.6 + (-4)t
-31.6 = -4t
t = 7.9 seconds.
To find the distance covered during this deceleration, we can use the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we get:
s = (31.6)(7.9) + (1/2)(-4)(7.9)^2
s = 249.44 - 197.68
s ≈ 51.76 meters.
3. Acceleration back to speed:
The acceleration of the automobile is 2.22 m/s^2. To find the time it takes for the automobile to reach a speed of 31.6 m/s, we can use the equation of motion: v = u + at.
The final velocity is 31.6 m/s, the initial velocity is 0 m/s (since the automobile is at rest), and the acceleration is 2.22 m/s^2. Plugging these values into the equation, we can solve for t:
31.6 = 0 + (2.22)t
t = 31.6 / 2.22
t ≈ 14.23 seconds.
To find the distance covered during this acceleration, we can again use the equation of motion: s = ut + (1/2)at^2.
Plugging in the values, we get:
s = (0)(14.23) + (1/2)(2.22)(14.23)^2
s = 0 + 223.41
s = 223.41 meters.
Now, let's add up the distances covered during the deceleration and acceleration segments:
Total distance = 51.76 + 223.41
Total distance ≈ 275.17 meters.
Therefore, when the automobile reaches a speed of 31.6 m/s, it is approximately 275.17 meters behind the train.