Question

A racing car starts from rest at the point A and moves with constant acceleration of 11ms-2 for 8s. The velocity it has reached after 8s is then maintained for T s. The racing car then decelerates from this velocity to 40ms-1, into a further 2s reaching point B. Given that the distance between A and B is 1404, find value of T

Answer 1

$$V=a_1t_1=(11)(8)=88\frac{m}{s}$$

$$d_1=0.5a_1t_1^2=0.5(11)(8)^2=352\ m$$

We have:

$$V^2-v^2=2a'd_3$$ $$a'=\frac{88-40}{2}=24\frac{m}{s^2}$$

$$88^2-40^2=2(24)d_3\to d_3=128\ m$$

The distance at second stage:

$$d_2=d-d_1-d_3=1404-352-128=924\ m$$

Thus,

$$T=\frac{d_2}{V}=\frac{924}{88}=10.5\ s$$