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 $t^{2}$ + 3t + 4)meters. The velocity when the acceleration is zero is

Moderate

A

3m/s
B

42m/s
C

–9m/s
D

–15m/s

Solution

S = ( $t^{3}$ - 6 $t^{2}$ + 3t + 4)
V = ds/dt ⇒ V = 3 $t^{2}$ -12t + 3
a = dv/dt ⇒ a = 6t - 12
When a = 0 ; t = 2sec
V = 3(2)2 - 12(2) + 3 = -9m/sec

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