Answer
(a) $4.3~s$
(b) $22~m$
(c) $10~m/s$
Work Step by Step
Let's find the position as function of time equation for both bicyclists. Assume the positive direction to be the direction the bicycles are traveling with the origin at the starting position of the bicyclist who is repairing a flat tire (the bicyclist 1). So, $x_{0_1}=0$. The initial time is the moment the bicyclist 1 starts. So, $v_{0_1}=0$.
Bicyclist 1: constant acceleration ($a=2.4~m/s^2$) ->
$x_1=x_{0_1}+v_{0_1}t+\frac{1}{2}at^2$
$x_1=0+0t+\frac{1}{2}(2.4~m/s^2)t^2$
$x_1=(1.2~m/s^2)t^2$
Bicyclist 2: constant speed ($3.5~m/s$). First, let's find the initial distance from bicyclist 2 to bicyclist 1:
$distance=(average~speed)(time~elapsed)=(3.5~m/s)(2~s)=7.0~m$
Now, when the bicyclist 1 starts, $x_{0_2}=7.0~m$:
$x_2=x_{0_2}+v_2t=7.0~m+(3.5~m/s)t$
(a) When the bicyclist 1 catches the bicyclist 2:
$x_1=x_2$
$(1.2~m/s^2)t^2=7.0~m+(3.5~m/s)t$
$(1.2~m/s^2)t^2-(3.5~m/s)t-7.0~m=0$
This is a quadratic equation for $t$.
$t=\frac{-(-3.5~m/s)±\sqrt {(-3.5~m/s)^2-4(1.2~m/s^2)(-7.0~m)}}{2(1.2~m/s^2)}$
$t_1=4.3~s$
$t_2=-1.4~s$ (this solution is not valid)
(b) $x_1=(1.2~m/s^2)t^2=(1.2~m/s^2)(4.3~s)^2=22~m$
(c) $v_1=v_{0_1}+at=0+(2.4~m/s^2)(4.3~s)=10~m/s$
