Question

A particle is moving along positive x axis. its position varies as x=t^3-3t^2+12t+20 where x is in metres and t is in seconds. find initial velocity of the particle

Answer 1

Hey There!!!

Here we are given position as a function of time:


$x = t^3-3t^2+12t+20$


Velocity is defined as the derivative of position with respect to time. So we have:

$v=\frac{dx}{dt} = 3t^2-6t+12$

Now, we want initial velocity. Whenever we have to find "initial" quantity, we have to take t = 0.

Thus, initial velocity means velocity at t=0.

At t=0, the velocity is:

$v = 0 + 0 + 12 \\ \\ \implies \boxed{v = 12 \, \, m / s}$


Thus, Initial Velocity is 12 m/s.



Hope it helps
Purva
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