Question

A fugitive tries to hop on a freight train traveling at a constant speed of 5.2 m/s . Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.3 m/s2 to his maximum speed of 6.1 m/s , which he then maintains. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

16 seconds

97 m

Explanation:

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{6.1-0}{1.3}\\\Rightarrow t=4.692\ s$

Train has moved a distance of $5.2\times 4.692=24.3984\ m$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 1.3\times 4.692^2\\\Rightarrow s=14.3096\ m$

Fugitive needs to run $24.3984-14.3096=10.0888\ m$ to make up

Relative speed of fugitive $6.1-5.2=0.9\ m/s$

Time taken

$t=\dfrac{10.0888}{0.9}\\\Rightarrow t=11.209\ s$

Time taken to catch the box = 4.692+11.209 = 15.901 seconds = 16 seconds

Distance traveled by the fugitive = $6.1\times 15.901=96.9961\approx 97\ m$