question_answer

A car accelerates from rest with a constant acceleration \[\alpha \] on a straight road. After gaining a velocity \[{{v}_{1}}\], the car moves with the same velocity for some-time. Then the car decelerated to rest with a retardation \[\beta \]. If the total distance covered by the car is equal to S, the total time taken for its motion is

A) \[\frac{S}{v}+\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]

B) \[\frac{S}{v}+\frac{v}{\alpha }+\frac{v}{\beta }\]

C) \[\left( \frac{v}{\alpha }+\frac{v}{\beta } \right)\]

D) \[\frac{S}{v}-\frac{v}{2}\left( \frac{v}{\alpha }+\frac{1}{\beta } \right)\]

Correct Answer: A

Solution :

From v = u + at, we have \[\text{v}=0+\alpha {{\text{t}}_{1}}\Rightarrow \frac{\text{v}}{\alpha }\] \[0=v-\beta {{t}_{3}}\Rightarrow {{t}_{3}}=\frac{\text{v}}{\beta }.\] \[\text{Now, S =}\frac{\text{1}}{\text{2}}\text{v}{{\text{t}}_{\text{1}}}\text{+v}{{\text{t}}_{\text{2}}}+\frac{1}{2}\text{v}{{\text{t}}_{\text{3}}}=\frac{{{\text{v}}^{\text{2}}}}{2\alpha }+\text{v}{{\text{t}}_{2}}+\frac{{{\text{v}}^{\text{2}}}}{2\beta }\] \[\Rightarrow {{\text{t}}_{\text{2}}}\text{=}\frac{\text{S}}{\text{v}}-\frac{\text{v}}{\text{2}}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right).\]\[\text{Hence, }{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}\text{+}{{\text{t}}_{\text{3}}}\text{=}\frac{\text{S}}{\text{v}}\text{+}\frac{\text{v}}{\text{2}}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]