A fugitive tries to hop on a freight train traveling at a constant speed of 4.8 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 m/s² to his maximum speed of 6.2 m/s.
How long does it take him to catch up to the empty box car?
The Correct Answer and Explanation is :
To solve this problem, we need to break it down into stages: the fugitive accelerates until he reaches his maximum speed, and then travels at that constant speed to catch up with the box car. Let’s go through each stage.
Step 1: Determine the time it takes for the fugitive to reach his maximum speed.
The fugitive starts from rest (initial velocity, ( u = 0 )) and accelerates at ( a = 1.4 \, \text{m/s}^2 ) to a maximum speed of ( v = 6.2 \, \text{m/s} ).
We can use the kinematic equation to find the time it takes for the fugitive to reach his maximum speed:
[
v = u + at
]
Substitute the known values:
[
6.2 = 0 + 1.4 \cdot t
]
Solving for ( t ):
[
t = \frac{6.2}{1.4} = 4.43 \, \text{seconds}
]
Step 2: Determine the distance traveled by the fugitive during this time.
Now, we can use another kinematic equation to find the distance ( d ) the fugitive covers while accelerating:
[
d = ut + \frac{1}{2} a t^2
]
Since the fugitive starts from rest (( u = 0 )):
[
d = \frac{1}{2} \cdot 1.4 \cdot (4.43)^2
]
[
d = 13.74 \, \text{meters}
]
Step 3: Determine the time it takes for the fugitive to catch up once he reaches his maximum speed.
Once the fugitive reaches his maximum speed of ( 6.2 \, \text{m/s} ), he continues to move at that speed, while the box car moves at a constant speed of ( 4.8 \, \text{m/s} ). The relative speed between the two is:
[
\text{Relative speed} = 6.2 – 4.8 = 1.4 \, \text{m/s}
]
The initial distance between the fugitive and the box car when the fugitive reaches his maximum speed is ( 13.74 \, \text{meters} ).
The time it takes the fugitive to close this gap is:
[
t = \frac{\text{distance}}{\text{relative speed}} = \frac{13.74}{1.4} = 9.82 \, \text{seconds}
]
Total time:
The total time to catch up is the sum of the acceleration time and the time to close the gap:
[
T = 4.43 + 9.82 = 14.25 \, \text{seconds}
]
Thus, the fugitive takes 14.25 seconds to catch up to the empty box car.
Now, I’ll generate an image to visualize the situation.
Here is the image showing the fugitive trying to catch up to the freight train. The scene illustrates the dynamics of the situation with the fugitive running next to the train. Let me know if you’d like further clarification!
