A particle starts its motion from rest and moves with constant acceleration for time $t_1$ and then it retards with constant rate for $t_2$.

Question

And comes to rest. Then the ratio of maximum speed and average speed during the complete motion will be

MY SOLUTION

Let acceleration be a.

Max speed

$$v=at_1$$ Also distance covered will be $$s=\frac{(2)(at_1^2)}{2}$$ $$=at_1^2$$ So average speed $$=\frac{at_1^2}{t_1+t_2}$$ Taking their ratio gives $$\frac{t_1+t_2}{t_1}$$ That’s as far as I got. But the answer is 2:1 and I have no idea on how to get there. Please help me proceed.

Thanks!

Answer 1

You have assumed that the magnitude of the acceleration and deceleration is equal, which is not given.

For the acceleration phase, $v_{max}=a_1t_1$ and distance covered is $\frac12a_1t_1^2$.

For the deceleration phase, $v^2-u^2=-a_1^2t_1^2=-2a_2s$ giving the distance covered as $\frac{a_1^2t_1^2}{2a_2}$. You also have $v=0=a_1t_1-a_2t_2$. Thus the average speed is $$\frac12\frac{a_1t_1^2+\frac{a_1^2t_1^2}{a_2}}{t_1+t_2}=\frac12\frac{a_1t_1^2+a_2t_2^2}{t_1+t_2}$$Taking the ratio$$\frac{v_{avg}}{v_{max}}=\frac12\frac{\frac{a_1t_1^2}{a_1t_1}+\frac{a_2t_2^2}{a_1t_1}}{t_1+t_2}=1/2$$