Question

A fugitive tries to hop on a freight train traveling at a constant speed of $\mathbf{5}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$. Just as an empty box car passes him, the fugitive starts from rest and accelerates at $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}}$to his maximum speed of $\mathbf{6}\mathbf{}{\mathbf{ms}}^{\mathbf{-}\mathbf{1}}$, which he then maintains. (a) How long does it take him to catch up to the empty box car? (b) What is the distance travelled to reach the box car?

Short answer

(a) The fugitive takes 7.14 s to catch the car.

(b) The fugitive travels a distance of 35.7 m before catching the car.

Step-by-step solution

Step 1. Meaning of uniform motion

The motion in which the velocity of the body does not change with time is called uniform motion.

Step 2. Data identification and assumption

Speed of the train,$V_{\mathrm{t}}=5{\mathrm{ms}}^{-1}$

Maximum speed of the fugitive,$v_{\circ }=6{\mathrm{ms}}^{-1}$

Acceleration of the fugitive,$a=1.4{\mathrm{ms}}^{-2}$

Let the time taken by the fugitive to reach the car be t, and the distance traveled before reaching the car be s.

Step 3. Calculation of the initial relative velocity

When the empty box car first passes the fugitive at rest, the speed of the train is $5{\mathrm{ms}}^{-1}$.

Then, the relative velocity of the fugitive with respect to the train is.

$\begin{array}{c}{v}_{\mathrm{rel}}=0-5\\ =-5{\mathrm{ms}}^{-1}\end{array}$

Step 4. Writing the second equation of motion

Let the relative distance between the fugitive and the train be d. Then by the second equation of motion, the value of d after time t is.

$d=\left(v_{\mathrm{rel}}\times t\right)+\frac{1}{2}a{t}^2$

Substituting the values in the above equation,

$d=-5t+\frac{1}{2}\times 1.4t^2$

When the fugitive catches the train, d becomes zero. Thus,

$0=-5t+\frac{1}{2}\times 1.4t^2$

Solving the above quadratic and neglecting t = 0,

$t=7.14\mathrm{s}$

Thus, the fugitive will take 7.14 s to catch the car.

Step 5. Calculation of the distance traveled before catching the box car

As the initial and final positions of the fugitive and the box car are the same (when the fugitive catches the box car), the distance traveled by both in time t is also the same. Thus, the distance traveled by the fugitive is.

$s=v_{\mathrm{T}}\times t$

Substituting the values in the above equation,

$\begin{array}{c}s=5\times 7.14\\ =35.7\mathrm{m}\end{array}$

Thus, the fugitive will travel a distance of 35.7 m before reaching the car.