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Question

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t32t), where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin

A

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B

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C

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D

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Solution

The correct option is (B)

Step 1, Given data

Velocity, v = 4t32t .

Step 2, Finding the equation

We know,

Acceleration a=dvdt

So, a=d(4t32t)dt

Or, a=dvdt=12t22

We also know that

v=dxdt

Or,dx=vdt

Now we can write

and x=vdt=(4t32t)dt=t4t2

According to the question when particle is at 2 m from the origin,

We can write

t4t2=2

Or,

t4t22=0

t4+t22t22=0

t2(t2+1)2(t2+1)=0

After solving this polynomial

(t22)(t2+1)=0t=2sec (only possible physical root)

Now this value of t will be substituted in the equation of acceleration

Acceleration at t=2 sec given by,

a=12t22=12×22=22m/s2

Hence the acceleration is 22 m/s2 .


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