Question

A car moves along a straight line whose equation is given by s=2t+3t^2-2t^3 where
sis in metres and tis in seconds. The velocity of the car at start will be m/sec​

Answer 1

Given :

▪ Displacement-time equation has been provided.

$\bigstar\:\underline{\boxed{\bf{\red{s=2t+3t^2-2t^3}}}}$

To Find :

▪ Velocity of the car at start (t = 0)

SoluTioN :

Instantaneous velocity is given by

$\Rightarrow\sf\:v=lim\:(\Delta t\to 0)\:\dfrac{\Delta s}{\Delta t}=\dfrac{ds}{dt}\\ \\ \Rightarrow\sf\:v=\dfrac{d(2t+3t^2-2t^3)}{dt}\\ \\ \Rightarrow\bf\:v=2+6t-6t^2\\ \\ \gray{\dag\sf\:Putting\:t=0\:we\:get,}\\ \\ \Rightarrow\sf\:v_0=2+6(0)-6(0)^2\\ \\ \Rightarrow\underline{\boxed{\bf{\blue{v_0=2\:ms^{-1}}}}}\:\orange{\bigstar}$

Answer 2

Given ,

" A car moves along a straight line whose equation is given by

s = 2t + 3t² - 2t³ where  s is in metres and tis in seconds "

We need to find the velocity of the car at starting point i.e., t = 0 .

We know that ,

$\bf Velocity , v = \dfrac{ds}{dt} \\\\\implies \bf v=\dfrac{d}{dt}(2t+3t^2-2t^3)\\\\\implies \bf v=2+6t-6t^2$

Now , velocity at t = 0 is ,

$\implies \bf v=2+6(0)-6(0)^2\\\\\implies \bf{\red{ v=2\ m/s}}$

So , velocity of the car at start is 2 m/s.