First slide

Displacement in nth sec

Question

The position x of a particle with respect to time t along X-axis is given by $x = 9 t^{2} - t^{3}$
where x is in metre and t in second. What will be the position of this particle when it achieves maximum speed along the + X direction?

Moderate

A

32 m
B

54 m
C

81 m
D

24 m

Solution

$x = 9 t^{2} - t^{3}$
$v = \frac{dx}{dt} = 18 t - 3 t^{2}$
$a = \frac{dv}{dt} = 18 - 6 t$
When $a = 0, v = maximum 18 = 6 t t = 3 \sec$
$x = 9 \times 3^{2} - {(3)}^{3} = 81 - 27 = 54 m$

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