Question

The motion of a particle along a straight line is described by the equation x = 4 – 12t + 3t3, where x is in metre and t is in second. Acceleration of the particle at t = 2 second will be

Answer 1

Given that, x = 4 - 12t + 3t³ where x is in metre and t is in second.

We know that,

v = dx/dt and a = dv/dt

So,

v = d(4 - 12t + 3t³)/dt

v = -12 + 3t²

Also,

a = d(-12 + 3t²)/dt

a = 3t

We have to find the acceleration of the particle at the time (t) = 2 second

Simply, substitute value of t = 2 sec in a = 3t. To find the value of acceleration.

→ a = 3(2)

→ a = 6

Therefore, of the particle at t = 2 second is 6 m/s².

Answer 2

Given:-

The motion of a particle along a straight line is described by the equation x = 4 - 12t + 3t³, where x is in metre and t is in second.

To find:-

Acceleration of the particle at t = 2 second.

Solution:-

We know that,

$\displaystyle{\frac{dx}{dt}=v }\\\\\textsf{or, velocity = rate of change in displacement,}\\\\\displaystyle{\frac{dv}{dt}=a }\\\\\textsf{or, acceleration is the rate of change in velocity.}$

v = d(4 - 12t + 3t³)/dt

⇒v = -12 + 3t²

Now, a = dv/dt

⇒a = (-12 + 3t²)/dt

⇒a = 3t

Acceleration when the time is 2 sec :

a = 3t

⇒a = 3*2

$\boxed{\implies a = 6 m/s\²}$