Answer to Question #242861 in Mechanics | Relativity for Lex

Question

Two riders initially at rest sit 5 m apart on a long, straight track, facing opposite directions. Starting at the same time, Rider A and B accelerates at 2.0 m/s2 and 3.0 m/s2, respectively. How far apart they will be at after 5.0 s?

Answer 1

First, we know that the riders are 5 m apart.

Then, since both riders start from the rest and accelerate Vi = 0 m/s, a₁ = 2.0 m/s² and a₂ = 3.0 m/s².

Now, the distance that each rider travels can be calculated with $d_i=a_it^2/2$ (this also comes from the fact that they both start to move from rest and Vi = 0 m/s).

The total distance between both riders will be the sum of the travel distance of each rider and the initial distance between them:

$d_{total}=d_{apart}+d_1+d_2 \\d_{total}=d_{apart}+(a_1t^2)/2+(a_2t^2)/2 \\d_{total}=d_{apart}+(\frac{a_1+a_2}{2} )t^2$

Now we substitute and find the total distance (at t = 5 s) as:

$\\d_{total}=(5\,m)+\bigg(\cfrac{(2.0+3.0)\frac{m}{\cancel{s^2}}}{2} \bigg)(5\,\cancel{s})^2 \\d_{total}=(5+(25)(2.5))\,m \\ d_{total}=(5+62.5)\,m=67.5\,m$

In conclusion, at t = 5 s we find that the two riders are 67.5 m apart from each other.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.