Question

Motion of a particle is given by x = 6t+ 3t^2 -2t^3m. Find its v at t=0s and t=1s. Also find its ‘a’ at t=1s.

Answer 1

Given :

Motion of particle is given by

$\sf\:x=6t+3t^2-2t^3$

To Find :

• Velocity at t = 0 & 1 sec
• Acceleration at t = 1 s

Solution :

We have,

$\sf\:x=6t+3t^2-2t^3$

Now differnatiate with respect to t

$\sf\dfrac{dx}{dt}=6+6t-6t^2$

$\sf\:v=6+6t-6t^2...(1)$

Put t = 0 sec ...in equation (1)

$\sf\:v=6ms^{-1}$

Put t = 1 sec .. in equation (1)

$\sf\:v=6ms^{-1}$

Thus , The Velocity of a particle at t = 0 sec is 6m/s and at t = 1 sec velocity is 6m/s

Now ,

$\sf\:v=6+6t-6t^2$

Differentiate with respect to t

$\sf\dfrac{dv}{dt}=6-12t$

$\sf\:a=6-12t...(2)$

Put t = 1 sec ...in equation (2)

$\sf\:a=-6ms^{-2}$

Therefore, acceleration of particle at t= 1 sec is -6m/s².

Answer 2

Given :

Motion of particle is given by

$\sf\:x=6t+3t^2-2t^3$

To Find :

1. Velocity at t = 0 and 1 sec
2. Acceleration at t = 1 s

Theory :

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

${\purple{\boxed{\large{\bold{Velocity=\frac{Distance}{Time\:interval}}}}}}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

• Acceleration

It is defined as the rate of change of velocity.

${\purple{\boxed{\large{\bold{Acceleration=\frac{Velocity}{Time\:interval}}}}}}$

In differential form:

$\sf\:Acceleration,a=\dfrac{dv}{dt}$

Solution :

Motion of particle given by

$\sf\:x=6t+3t^2-2t^3$

Part -1

We have to find the velocity of the particle.

$\sf\:x=6t+3t^2-2t^3$

Now differnatiate with respect to t

$\sf\dfrac{dx}{dt}=6+6t-6t^2$

$\sf\:v=6+6t-6t^2$

If t = 0 sec

Then, $\sf\:v=6ms^{-1}$

if t= 1 sec

Then ,$\sf\:v=6ms^{-1}$

Hence ,The Velocity of a particle

• At t = 0 sec is 6m/s
• And at t = 1 sec is 6m/s

Part -2

We have to find the acceleration of the particle.

$\sf\:v=6+6t-6t^2$

Now ,Differentiate with respect to t

$\sf\dfrac{dv}{dt}=6-12t$

$\sf\:a=6-12t$

If t = 1 sec

Then, $\sf\:a=-6ms^{-2}$

Hence,The acceleration of particle at t= 1 sec is -6m/s².

$\rule{200}2$

More information about topic

• Both Velocity and Acceleration are vector quantities.
• The velocity of an object can be positive, zero and negative.
• SI unit of Velocity is m/s
• SI unit of Acceleration is m/s²
• Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$
• Dimension of Acceleration: $\sf\:[M^0LT{}^{-2}]$