Question

→ Find velocity at, t = 5 secMotion of particle is given by$\sf\:x(t)=4t^2+8t$m​

Answer 1

$\:\:\:\: \large\underline {\sf{To\:Find\::}}$

• Velocity at t = 5 sec

$\:\:\:\: \large\underline {\sf{Solution\::}}$

We know that,

${\underline {\boxed {\small {\sf { Velocity (v)=\dfrac{Distance (d)}{Time(t)} } }}}}$

• Motion of particle is given by

$\sf\:x(t)=4t^2+8t$m

In differential form:

• $\sf\:Velocity,V=\dfrac{dx}{dt}$

Now, find the velocity of the particle at t = 5 sec

$\sf\:x=4t^2+8t$

Differentiate with respect to t :-

• $\sf\dfrac{dx}{dt}=\dfrac{d(4t^2)}{dt}+\dfrac{d(8t)}{dt}$

We know that,

• $\sf\dfrac{d(x^n)}{dx}=nx^{n-1}$

$:\implies\sf\:v=8t+8$

If t = 5sec

Then,

$:\implies\sf\:v=8\times5+8$

$:\implies \sf\:v=40+8$

$:\implies\sf\:v=48ms^{-1}$

$:\implies{\boxed{\sf{\purple{ 48ms^{-1}}}}}$

$\Large{\underbrace{\sf{\red{Know\:More:}}}}$

• The rate of change of displacement of a particle with time is called velocity of the particle.
• Velocity is a vector quantity.
• The velocity of an object can be positive, zero and negative.
• SI unit of Velocity is m/s
• Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$

________________________

Answer 2

V=48m s^-1 using differentiation