First slide

Instantaneous velocity

Question

A particle moves along a straight line OX. At a time t (in seconds), the distance . $x = 40 + 12 t - t^{3}$ Then the displacement of the particle before coming to rest is

Moderate

A

24 m
B

40 m
C

56 m
D

16 m

Solution

Distance travelled by the particle is $x = 40 + 12 t - t^{3}$
We know that, speed is defined as the rate of change of distance, i.e. $v = \frac{d x}{d t}$
$∴$ $v = \frac{d}{d t} (40 + 12 t - t^{3}) = 0 + 12 - 3 t^{2}$
But final velocity, v = 0
$∴$ $𝟚 - 3 t^{2} = 0 \Rightarrow t^{2} = \frac{12}{3} = 4$
$\Rightarrow$ t = 2s
Hence, distance travelled by the particle before coming to rest is given by
x = 40 + 12(2) - (2)³ = 40 + 24 - 8 = 64 - 8 = 56 m

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