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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?
By BYJU'S Exam Prep
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.