First slide

Instantaneous acceleration

Question

The velocity of a particle moving along X axis changes with position as per the equation given by $v^{2} = (x^{2} + 2 x - 1)$ . Find the instantaneous tangential acceleration of the particle at x = 1m.

Easy

A

1 $m s^{- 2}$
B

2 $m s^{- 2}$
C

3 $m s^{- 2}$
D

4 $m s^{- 2}$

Solution

$v^{2} = x^{2} + 2 x - 1, a = v \frac{d v}{d x}$
Now differentiating the first equation with respect to x is
$2 v \frac{d v}{d x} = 2 x + 2, v \frac{d v}{d x} = x + 1$
So at x = 1, a = 2 $m s^{- 2}$

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