# A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), 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?

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Updated on: September 25th, 2023

(a) 28 m/s2

(b) 22 m/s2

(c) 12 m/s2

(d) 10 m/s2

The acceleration of the particle, when it is 2 m from the origin is 22 m/s2. Steps to find the acceleration of the particle, when it is 2 m from the origin:

Step 1: Given that:

Velocity, v = 4t3 – 2t

Step 2: Now we have to find the equation:

We know that Acceleration a = dv/dt

Substituting the values we get:

So, a = d (4t3 – 2t)/ dt

a = dv/dt = 12t2 – 2

We also know that:

v = dx/dt

Or ∫dx = ∫vdt

The above equation can be written as:

and x = ∫vdt = ∫(4t3 – 2t)dt = t4 – t2

When a particle is 2 meters from its source, the question states,

We can write:

t4 – t2 = 2

Or

t4 – t2 – 2 = 0

t4 + t2 – 2t2 – 2 = 0

t2 (t2 + 1) – 2 (t2 + 1) = 0

After solving the above polynomial we get:

(t2 – 2) (t2 + 1) = 0

t = √2 sec (only possible physical root)

The equation for acceleration will now substitute this number for t.

Acceleration at t = √2 sec given by:

a = 12t2 – 2 = 12 x 2 – 2 = 22 m/s2

therefore the acceleration is 22 m/s2.

Summary:

## A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), 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) 28 m/s2 (b) 22 m/s2 (c) 12 m/s2 (d) 10 m/s2

A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3 – 2t), where t is in sec and velocity in m/s. The acceleration of the particle, when it is 2 m from the origin is 22 m/s2.

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