#### Answer

(a) The time required to catch the box car is 26 s.
(b) The distance traveled to reach the box car is 143 m.

#### Work Step by Step

$t_1 = \frac{v-v_0}{a} = \frac{6.0 ~m/s}{1.4 ~m/s^2} = 4.3 ~s$
It takes 4.3 s to accelerate up to a speed of 6.0 m/s.
$x = v\cdot t_1 = (5.0 ~m/s)(4.3 ~s) = 22 ~m$
In 4.3 s, the box car moves a distance of 22 m.
The fugitive is moving at 6.0 m/s and the box car is moving at a speed of 5.0 m/s. Therefore, the fugitive will catch the box car at a rate of 1.0 m/s.
$t_2 = \frac{x}{v} = \frac{22 ~m}{1.0 ~m/s} = 22 ~s$
(a) The total time to catch the box car is $t_1 + t_2$.
$t_1+ t_2 = 4.3 ~s + 22 ~s = 26 ~s$
The time required to catch the box car is 26 s.
(b) Let $x_1$ be the distance traveled during the acceleration period.
$x_1 = \frac{1}{2}at^2 = \frac{1}{2}(1.4 ~m/s^2)(4.3 ~s)^2 = 13 ~m$
Let $x_2$ be the distance traveled while moving at a constant speed of 6.0 m/s.
$x_2 = vt = (6.0 ~m/s)(22 ~s) = 130 ~m$
The total distance is $x_1 + x_2$.
$x_1 + x_2 = 13 ~m + 130 ~m = 143 ~m$
The distance traveled to reach the box car is 143 m.