A body starts from rest and travels a distance $x$ with uniform acceleration, then it travels a distance $2 \mathrm{x}$ with uniform speed, finally it travels a distance $3 x$ with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the…

Question

A body starts from rest and travels a distance $x$ with uniform acceleration, then it travels a distance $2 \mathrm{x}$ with uniform speed, finally it travels a distance $3 x$ with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the ratio of its average velocity to maximum velocity is

Aicharya · Accepted Answer

Area of $(v-t)$ graph=displacement For uniform acceleration $s=\frac{1}{2} \times t_1 \times v_{\max }$ $t_1=\frac{2 s}{v_{\max }} \ldots .. (1)$ For uniform speed $2 s=t_2 \times v_{\max }$ $t_2=\frac{2 s}{v_{\max }} \ldots . . (2) $ For uniform retardation $3 s=\frac{1}{2} \times t_3 \times v_{\max }$ $t_3=\frac{6 s}{v_{\max }} \ldots . . (3) $ $\begin{aligned} & \left|\vec{v}_{\text {avg }}\right|=\frac{\text { Displacement }}{\text { Total time }}=\frac{s+2 s+3 s}{t_1+t_2+t_3} \ & \frac{6 s}{\frac{2 s}{v_{\max }}+\frac{2 s}{v_{\max }}+\frac{6 s}{v_{\max }}}=\frac{6 s}{\frac{10 s}{v_{\max }}}=\frac{6 v_{\max }}{10} \ & \left|\vec{v}_{\text {avg }}\right|=\frac{6}{10} \times v_{\max } \text { then } \frac{\left|\vec{v}_{\max }\right|}{v_{\max }}=\frac{3}{5} \end{aligned}$

A body starts from rest and travels a distance $x$ with uniform acceleration, then it travels a distance $2…

Solution