Question

A particle starts from rest and travels a distance $x$ m with uniform acceleration, then moves uniformly a distance $2x$ m and finally comes to rest after moving further $5x$ m with uniform retardation. The ratio of maximum speed to average speed is:

Answer 1

It is clear that maximum velocity is at $A$ and from $A$ to $-B$ is $\left(V\right)$

area under curve is displacement

$\frac{v{t}_1}{2}=x$,$v{t}_2=2x$ and $\frac{v{t}_3}{2}=5x$

$(t_1+t_2+t_3)=\frac{2x}{v}+\frac{2x}{v}+\frac{10x}{v}=\frac{14x}{v}$

average velocity $\overline{V}=(5x+2x+x)/(t_1+t_2+t_3)\overline{V}=8x/\frac{14x}{v}$

$\overline{V}=\frac{8}{14}V=\frac{4}{7}V\frac{V}{\overline{V}}=\frac{7}{4}$