A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m / s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at 4.0 m / s^2 to a maximum speed of 8.0 m / s. (a) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car?
Answer
Kinematics Analysis: Fugitive Catching a Moving Boxcar
This physics problem involves a classic relative motion and acceleration scenario. A fugitive starts running to catch up with a freight train that is already in motion at a constant speed. We’ll break it down using kinematic equations.
Given:
- Speed of train (vtrain) = 6.0 m/s
- Fugitive starts from rest: initial velocity (u) = 0
- Fugitive acceleration (a) = 4.0 m/s²
- Maximum speed of fugitive (vmax) = 8.0 m/s
Step 1: Time to Reach Max Speed
This is the time it takes the fugitive to reach his top speed.
Step 2: Distance During Acceleration Phase
Use the formula for displacement during uniform acceleration:
Step 3: Distance Covered by Train in 2 Seconds
So after 2 seconds, the train is 12.0 meters away, while the fugitive has only traveled 8.0 meters. He still needs to cover the remaining gap of:
Step 4: Time to Close the Remaining Distance at Constant Speed
After reaching his max speed of 8.0 m/s, the fugitive maintains that speed to catch the train, which is still moving at 6.0 m/s. Their relative speed is:
Time to close the 4-meter gap:
Step 5: Total Distance Traveled by Fugitive
Distance during constant-speed phase:
Add acceleration phase distance (8.0 m):
This is a textbook example of breaking down motion into segments: acceleration and uniform motion. Always compare relative speeds when chasing or catching up in motion problems!