Kinematics Question 380
Question: 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 it’s motion is
Options:
A) $ \frac{S}{v}+\frac{v}{2}( \frac{1}{\alpha }+\frac{1}{\beta } ) $
B) $ \frac{S}{v}+\frac{v}{\alpha }+\frac{v}{\beta } $
C) $ ( \frac{v}{\alpha }+\frac{v}{\beta } ) $
D) $ \frac{S}{v}-\frac{v}{2}( \frac{v}{\alpha }+\frac{1}{\beta } ) $
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Answer:
Correct Answer: A
Solution:
From v = u + at, we have $ v=0+\alpha {{t} _{1}}\Rightarrow \frac{v}{\alpha } $
$ 0=v-\beta t _{3}\Rightarrow t _{3}=\frac{v}{\beta }. $
$ \text{Now, S =}\frac{1}{2}v{{t} _{1}}\text{+v}{{t} _{2}}+\frac{1}{2}v{{t} _{3}}=\frac{{{v}^{2}}}{2\alpha }+v{{t} _{2}}+\frac{{{v}^{2}}}{2\beta } $
$ \Rightarrow {{t} _{2}}\text{=}\frac{S}{v}-\frac{v}{2}( \frac{1}{\alpha }+\frac{1}{\beta } ). $
$ \text{Hence, }{{t} _{1}}\text{+}{{t} _{2}}\text{+}{{t} _{3}}\text{=}\frac{S}{v}\text{+}\frac{v}{2}( \frac{1}{\alpha }+\frac{1}{\beta } ) $
BETA
