First slide

Instantaneous acceleration

Question

A particle moves along a straight line such that its displacement at any time t is given by
$s = t^{3} - 6 t^{2} + 3 t + 4$
The velocity when its acceleration is zero, is

Easy

A

$3 {ms}^{- 1}$
B

$- 12 {ms}^{- 1}$
C

$42 {ms}^{- 1}$
D

$- 9 {ms}^{- 1}$

Solution

$x = t^{3} - 6 t^{2} + 3 t + 4$
Velocity $v = \frac{dx}{dt} = 3 t^{2} - 6 \times 2 t + 3 \times 1 + 0$
= $3 t^{2} - 12 t + 3$
Acceleration a = $\frac{dv}{dt} = 3 \times 2 t - 12 \times 1 + 0 = 6 t - 12$
Acceleration is zero at time t given by 6t-12 = 0
$\Rightarrow t = 2 seconds$
So, velocity v at t = 2 seconds is
v = ${(3 t^{2} - 12 t + 3)}_{t = 2 s}$
$= 3 \times {(2)}^{2} - 12 \times 2 + 3 = - 9 m / s$

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