Question

A body moves with an initial velocity of 30ms -1 and accelerates uniformly until it attains the velocity 80ms-1. It then continue at that velocity for some time and decelerates uniformly to rest. The total time taken for the journey is 40 and the total distance traveled is 2550 km. If the time spent accelerating is half that of traveling at constant velocity.calculate the acceleration 

Answer 1

Let:

$t_1$ is a time of acceleration;

$t_2$ is a time of uniform motion;

$t_3$ is a time of deceleration.

We get

$$t_1+t_2+t_3=6 \;\rm hr$$

$$t_1+t_3=t_2/2$$

Thus,

$$t_2=12/3\:\rm hr=14400\: s$$

$$t_3=14400\:{\rm s}-t_1$$

The total distance

$$d=d_1+d_2+d_3$$

$$d=\frac{v_1+v_2}{2}t_1+v_2t_2+\frac{v_2+0}{2}t_3$$

$$2550000\:{\rm m}=\frac{30\: {\rm m/s}+80\:{\rm m/s}}{2}*t_1+80\:{\rm m/s}*14400\:{\rm s}$$

$$\\+\frac{80\:\rm m/s+0}{2}*(14400\:{\rm s}-t_1)$$ $$822000=15t_1$$

We obtain

$$t_1=54800\:\rm s$$

The acceleration

$$a=\frac{v_2-v_1}{t_1}=\frac{80-30}{54800}=0.0009\:\rm m/s^2$$