Hello Fallin' staR,
We will let $a(t)$ be acceleration, $v(t)$ be velocity, and $s(t)$ be position, and use a subscript of $\text{M}$ for the motorbike and a subscript of $\text{C}$ for the car.
For the motorbike, during the time its acceleration is greater than zero, that is, when its velocity is changing, we may state:
$$\int_0^{v(t)}\,du=\frac{5}{2}\int_0^t\,dw$$
$$v(t)=\frac{5}{2}t$$
Now, we need to find at what value of $t$ we have $v(t)=20$, thus:
$$20=\frac{5}{2}t$$
Solve for $t$:
$$t=8$$
And so we may now give the velocity of the motorbike as:
$$v_{\text{M}}(t)=\begin{cases}\frac{5}{2}t & t<8\\ 20 & 8\le t \\ \end{cases}$$
For the car, during the time its acceleration is greater than zero, that is, when its velocity is changing, we may state:
$$\int_0^{v(t)}\,du=\frac{3}{2}\int_0^t\,dw$$
$$v(t)=\frac{3}{2}t$$
Now, we need to find at what value of $t$ we have $v(t)=30$, thus:
$$30=\frac{3}{2}t$$
Solve for $t$:
$$t=20$$
And so we may now give the velocity of the car as:
$$v_{\text{C}}(t)=\begin{cases}\frac{3}{2}t & t<20\\ 30 & 20\le t \\ \end{cases}$$
Here is a plot of the two velocity functions on the same axes:
View attachment 1234
To generate this graph, I used the command:
piecewise[{{(5/2)t,0<=t<8},{20,8<=t}}],piecewise[{{(3/2)t,0<=t<20},{30,20<=t}}] where t=0 to 30
at
Wolfram|Alpha: Computational Knowledge Engine
Now, to find the position function of the motorbike, we may use:
On the interval $0\le t<8$:
$$\int_0^{s(t)}\,du=\frac{5}{2}\int_0^{t} w\,dw$$
$$s(t)=\frac{5}{4}t^2$$
We will need to know that:
$$s(8)=80$$
On the interval $8\le t$:
$$\int_{s(8)}^{s(t)}\,du=20\int_8^t \,dw$$
$$s(t)=20t-80$$
Thus, we may state:
$$s_{\text{M}}(t)=\begin{cases}\frac{5}{4}t^2 & t<8\\ 20t-80 & 8\le t \\ \end{cases}$$
Now, to find the position function of the car, we may use:
On the interval $0\le t<20$:
$$\int_0^{s(t)}\,du=\frac{3}{2}\int_0^t w\,dw$$
$$s(t)=\frac{3}{4}t^2$$
We will need to know that:
$$s(20)=300$$
On the interval $20\le t$:
$$\int_{s(20)}^{s(t)}\,du=30\int_{20}^t \,dw$$
$$s(t)=30t-300$$
Thus, we may state:
$$s_{\text{C}}(t)=\begin{cases}\frac{3}{4}t^2 & t<20\\ 30t-300 & 20\le t \\ \end{cases}$$
Using the command:
piecewise[{{(5/4)t^2,0<=t<8},{20t-80,8<=t}}],piecewise[{{(3/4)t^2,0<=t<20},{30t-300,20<=t}}] where t=0 to 30
we obtain the plot:
View attachment 1235
Now we may answer the two questions:
(a) after what time will the motorbike and the car again be side by side?
From the graph, we see we want to equate the two linear portions of the position functions:
$$20t-80=30t-300$$
$$10t=220$$
$$t=22\text{ s}$$
Thus, we find that the motorbike and the car are again side by side 22 seconds after the traffic light turns green.
(b) what is the greatest distance that the motorbike is in front of the car?
The graph shows us that this will occur on the interval $8<t<20$. Hence we want to maximize the function:
$$f(t)=\left(20t-80 \right)-\left(\frac{3}{4}t^2 \right)=-\frac{3}{4}t^2+20t-80$$
Completing the square, we find:
$$f(t)=-\frac{3}{4}\left(t-\frac{40}{3} \right)^2+\frac{160}{3}$$
Hence:
$$f_{\max}=\frac{160}{3}\text{ m}$$
And so we find the greatest distance that the motorbike is in front of the car to be about 53.3 m.
