A body starting from rest, accelerates at a constant rate a m / s… - Vidyadip

MCQ

A body starting from rest, accelerates at a constant rate a $m / s ^2$ for some time, after which it decelerates at a constant rate b $m / s ^2$ to come to rest finally. If the total time elapsed is t s the maximum velocity attained by the body is given by

✓
$\frac{a b}{a+b} t m / s$
B
$\frac{a b}{a-b} t m / s$
C
$\frac{2 a b}{a+b} t m / s$
D
$\frac{2 a b}{a-b} t m / s$

✓

Answer

Correct option: A.

$\frac{a b}{a+b} t m / s$

(A)
Total time of motion $= t$
Duration of acceleration $=t^{\prime}$
Duration of deceleration $=t-t^{\prime}$
Given u = 0, a = constant acceleration and
b = constant deceleration.
$v = 0 + at ^{\prime}$
Also $0= v - b \left( t - t ^{\prime}\right)$
$\therefore \quad v = at ^{\prime}$
From (ii), $- v =- bt + bt ^{\prime}$
$\Rightarrow- at ^{\prime}=- bt + bt ^{\prime}$
$\Rightarrow( a + b ) t ^{\prime}= bt \Rightarrow t ^{\prime}=\frac{ b }{( a + b )} t$
But $v = a t ^{\prime}$
$\therefore \quad$ Maximum velocity attained $= at ^{\prime}$
$\Rightarrow v =\frac{ ab }{( a + b )} t m / s$

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