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Question

A point initially at rest moves along X-axis. Its acceleration varies with time as a=(6t+5)ms-2. If it starts from origin, the distance covered in 2s is


A

20m

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B

18m

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C

16m

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D

25m

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Solution

The correct option is B

18m


Step 1: Given data

Acceleration of the point, a=(6t+5)ms-2

Initial velocity of particle, u=0ms-1

Time taken by the particle, t=2s

Step 2: Find the velocity of the particle

As, acceleration, a=dvdt [v is the velocity]

Therefore, dvdt=6t+5,

dv=(6t+5)dt

On integrating it,

0vdv=0t(6t+5)dtv=6t22+5t+cv=3t2+5t+c

At t=0, v will be also 0. So,

0=0+0+cc=0

Therefore,

Now, v=3t2+5tdsdt=3t2+5tds=(3t2+5t)dtds=02(3t2+5t)dts=t3+(52)t202s=8+10=18m [ Asv=dsdt, s is distance]

Hence, option (B) is correct.


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