I have tried to solve this problem by adding the sum of the displacements during acceleration, constant velocity and deceleration, but it does not work out.
Question:
A car accelerates from rest to $20~\text{m/s}$ in $12$ seconds ($a =5/3~\text{ms}^{-2}$), it travels at $20~\text{m/s}$ for $40$ seconds, then retardation occurs from $20~\text{m/s}$ to rest in $8$ seconds ($a = -2.5~\text{ms}^{-2}$). As the car accelerates an RC car, moving parallel to the car, is moving at $14~\text{m/s}$. When will overtaking occur and what will the distance be? The RC car passes the car just as it starts to accelerate.
I can do this without a problem if acceleration is a constant. Is there a differential equation I can use to compute this as that is my better area or must I stick with the SUVAT equations?
Again, if I could be pointed in the right direction that way I can learn.
Best Answer
Since the acceleration changes discontinuously, there is no one way to solve it in one step.
One simplification is to switch to the frame of reference of the RC car. You can do this simply by making the initial velocity of the starting/stopping car $-14$ m/s. The question then reduces to: When is the displacement of the moving car zero again?
if you use SAVTU on the first acceleration, you can find out if $S=0$ in less than 12 seconds. If not, you can find out $S$ after 12 seconds, and then see if you get back to $S=0$ at constant velocity of $6$ m/s in less than 40 seconds, and so on...
Explicit solution added;
First, Setting $s=0$ (car catches RC during first acceleration), $a = \frac{5}{3}$, and $u = -14$, and solving for t, we get t = 16.8 sec
It would take 16.8 sec of constant acceleration for the car to catch the RC, which we don't have!
So, finding s at the end of the first acceleration, if $a=\frac{5}{3}$, $u = -14$, $t=12$, we get $$s=-14 \times 12+\frac{1}{2}\times\frac{5}{3}\times 12^2=-48$$The car is 48 m behind the RC at the end of the firat acceleration.
The car is catching the RC at $6 $ m/s, and will make up the 48 m in 8 seconds of constant velocity, which we do have.
Total time is 20 seconds. The RC will cover 280 m, and so will the accelerating car...