Question

Two cars P & Q start from a point at the same time in a straight line and their positions are represented by x1 = at+bt^2 & x2=ft-bt^2 . At what time do the cars have the same velocity?

Answer 1

Given:

For car P , the Displacement vs Time equation is :

x1 = at + bt²

For car Q , the Displacement vs Time equation is :

x2 = ft - bt²

To find:

Time at which both the cars will have same velocity.

Concept:

Any Displacement vs Time function can be differentiated (In Calculus) with respect to time in order to get a velocity function.

The velocity function for both the cars can be equated to get the required time.

Calculation:

For car P :

$x1 = at + b{t}^{2}$

$= > v1 = \dfrac{d(x1)}{dt}$

$= > v1 = \dfrac{d(at + b {t}^{2} )}{dt}$

$= > v1 = \dfrac{d(at)}{dt} + \dfrac{d(b {t}^{2} )}{dt}$

$= > v1 = a + 2bt$

For car Q:

$x2 = ft - b{t}^{2}$

$= > v2 = \dfrac{d(x2)}{dt}$

$= > v2 = \dfrac{d(ft - b {t}^{2} )}{dt}$

$= > v2 = \dfrac{d(ft)}{dt} - \dfrac{d(b {t}^{2} )}{dt}$

$= > v2 = f - 2bt$

Now , equating the two velocity functions , we get :

$\therefore \: v1 = v2$

$= > a + 2bt = f - 2bt$

$= > 4bt = f - a$

$= > t = \dfrac{f - a}{4b}$

So the time in which the velocity of the two cars will be equal is :

$\boxed{ \red{ \huge{ \bold{t = \dfrac{f - a}{4b} }}}}$