Answer to Question #92499 in Mechanics | Relativity for Sana

Question

A car starting from rest accelerates at the rate of f through distance S then continues at constant speed for tome t and decelerates at the rate f/2 to come to rest .if total distance travelled is 15s then S

Answer 1

The constant speed is

$$v^2=2fs$$

$$v=\sqrt{2fs}$$

Distance for the second period is

$$D=vt=\sqrt{2fs}t$$

For the third period:

$$v^2=2(0.5f)d$$ $$2fs=fd$$

$$d=2s$$

Total distance travelled is 

$$15s=s+\sqrt{2fs}t+2s$$

$$\sqrt{2fs}t=12s$$

Thus,

$$s=\frac{ft^2}{72}$$