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
The Correct Answer and Explanation is:
Let’s break this problem into parts to solve it:
Part (a): Time to Catch Up
We need to find the time it takes for the fugitive to catch up with the box car.
Step 1: Calculate the time it takes for the fugitive to reach his maximum speed.
The fugitive starts from rest, so his initial speed is 0. His acceleration is 4.0 m/s24.0 \, \text{m/s}^24.0m/s2 and his maximum speed is 8.0 m/s8.0 \, \text{m/s}8.0m/s. We can use the equation for velocity under constant acceleration:v=u+atv = u + atv=u+at
where:
- v=8.0 m/sv = 8.0 \, \text{m/s}v=8.0m/s (final velocity),
- u=0 m/su = 0 \, \text{m/s}u=0m/s (initial velocity),
- a=4.0 m/s2a = 4.0 \, \text{m/s}^2a=4.0m/s2 (acceleration),
- ttt is the time to reach maximum speed.
Substituting the known values:8.0=0+(4.0×t)8.0 = 0 + (4.0 \times t)8.0=0+(4.0×t)t=8.04.0=2.0 secondst = \frac{8.0}{4.0} = 2.0 \, \text{seconds}t=4.08.0=2.0seconds
So, it takes the fugitive 2.0 seconds to reach a speed of 8.0 m/s.
Step 2: Calculate the distance the fugitive travels during this acceleration.
Now that we know the fugitive’s acceleration time, we can find the distance traveled while accelerating using the kinematic equation:d=ut+12at2d = ut + \frac{1}{2} a t^2d=ut+21at2
Since u=0u = 0u=0, the equation simplifies to:d=12at2=12(4.0)(2.0)2=8.0 metersd = \frac{1}{2} a t^2 = \frac{1}{2} (4.0) (2.0)^2 = 8.0 \, \text{meters}d=21at2=21(4.0)(2.0)2=8.0meters
So, the fugitive travels 8.0 meters during the first 2.0 seconds.
Step 3: Determine the relative speed.
Once the fugitive reaches his maximum speed of 8.0 m/s, he will move at the same speed as the box car, which is also moving at 6.0 m/s. Therefore, the relative speed between the fugitive and the box car is:vrelative=8.0 m/s−6.0 m/s=2.0 m/sv_{\text{relative}} = 8.0 \, \text{m/s} – 6.0 \, \text{m/s} = 2.0 \, \text{m/s}vrelative=8.0m/s−6.0m/s=2.0m/s
Step 4: Find the time it takes to close the gap.
Initially, the fugitive is 8.0 meters behind the box car, and he is closing this gap at a rate of 2.0 m/s. The time to close the gap is:tgap=distancerelative speed=8.02.0=4.0 secondst_{\text{gap}} = \frac{\text{distance}}{\text{relative speed}} = \frac{8.0}{2.0} = 4.0 \, \text{seconds}tgap=relative speeddistance=2.08.0=4.0seconds
So, it takes 4.0 seconds to catch up after reaching maximum speed.
Total Time:
The total time to catch up is the time to accelerate (2.0 seconds) plus the time to close the gap (4.0 seconds):ttotal=2.0+4.0=6.0 secondst_{\text{total}} = 2.0 + 4.0 = 6.0 \, \text{seconds}ttotal=2.0+4.0=6.0seconds
Thus, the fugitive takes 6.0 seconds to catch up with the box car.
Part (b): Distance Traveled to Catch the Box Car
Step 1: Distance traveled during acceleration.
As calculated earlier, the fugitive travels 8.0 meters during the first 2.0 seconds while accelerating.
Step 2: Distance traveled while closing the gap.
After the fugitive reaches his maximum speed of 8.0 m/s, he will be closing the gap at a relative speed of 2.0 m/s. The time to close the gap is 4.0 seconds, so the distance he travels during this time is:dgap=relative speed×time=2.0 m/s×4.0 seconds=8.0 metersd_{\text{gap}} = \text{relative speed} \times \text{time} = 2.0 \, \text{m/s} \times 4.0 \, \text{seconds} = 8.0 \, \text{meters}dgap=relative speed×time=2.0m/s×4.0seconds=8.0meters
Total Distance:
The total distance traveled by the fugitive is the sum of the distance during acceleration and the distance while closing the gap:dtotal=8.0 meters+8.0 meters=16.0 metersd_{\text{total}} = 8.0 \, \text{meters} + 8.0 \, \text{meters} = 16.0 \, \text{meters}dtotal=8.0meters+8.0meters=16.0meters
Thus, the fugitive travels a total distance of 16.0 meters to catch up with the box car.
Final Answers:
(a) The time it takes the fugitive to catch up to the box car is 6.0 seconds.
(b) The total distance traveled to reach the box car is 16.0 meters.
