First slide

Instantaneous acceleration

Question

The position x of a particle varies with time t as x = a $t^{2}$ – b $t^{3}$ . The acceleration of the particle will be zero at time t equal to

Moderate

A

a/b
B

2a/3b
C

a/3b
D

Zero

Solution

Position x = a $t^{2}$ - b $t^{3}$
Velocity =
$\frac{d x}{d t} = \frac{d}{d t} (a t^{2} - b t^{3})$
Velocity = 2ta - 3b $t^{2}$
Acceleration =
$\frac{d v}{d t} = \frac{d}{d t} (2 a t - 3 b t^{2})$

Acceleration = 2a - 6bt = 0
2a = 6bt ⇒t = a/3b

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