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The position x of a particle with respect to time t along x-axis is given by x=9t2−t3 where x is in metres and t is in seconds. What will be the position of this particle when it achieves maximum speed along the + x direction?

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Solution

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x=9t2−t3.....(I)

We know that,

v=dxdt

The maximum speed is

v=18t−3t2

v maximum, at dvdt=0

dvdt=18−6t

The time will be

0=18−6t

t=3 sec

The position of the particle at 3 sec

Put the value of t in equation (I)

x=9×9−27

x=81−27

x=54 m

The position of the particle will be 54 m.

Hence, Option A is correct

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