Question

the position x of a particle varies with time t as x=at2-bt3.What is the velocity of particle at t=1sec​

Answer 1

Answer :

Position equation of particle has been provided.

• x = at² - bt³

We have to find velocity of particle at t = 1s$.$

Instantaneous velocity of moving body is given by ☃

• v = (lim t➝0) Δx/Δt = dx/dt

⇒ v = dx/dt

⇒ v = d(at² - bt³)/dt

⇒ v = 2at - 3bt²

Putting t = 1, we get

⇒ v = 2a(1) - 3b(1)²

⇒ v = 2a - 3b

Answer 2

Given ,

The position of a particle varies with time as x = at² - bt³

We know that , the instanteous velocity of a particle is given by

$\boxed{ \sf{v = \frac{dx}{dt} }}$

Thus ,

$\sf \hookrightarrow \frac{dx}{dt} = \frac{d \{a {t}^{2} - b {r}^{3} \} }{dt} \\ \\ \sf \hookrightarrow \frac{dx}{dt} =2at - 3b {t}^{2}$

At t = 1 sec , we get

$\sf \hookrightarrow \frac{dx}{dt} = 2a(1) - 3b {(1)}^{2} \\ \\ \sf \hookrightarrow \frac{dx}{dt} =2a - 3b$

Therefore ,

The velocity of particle at 1 sec is 2a - 3b