Question

Two cars P and Q start from a point at the same time in a straight line and their positions are represented by xp(t)=at+bt² and xq(t)=ft-t². At what time do the cars have the same velocity?

Answer 1

this may help u..........

Answer 2

Answer: At time $t = \dfrac{f-a}{2(1+b)}$ the cars will have same velocity Explanation:

Given that,

Position of P = $X_{p}(t)=at+bt^2$

Position of Q= $X_{q}(t)=ft+ft^2$

We know that,

Velocity of P= $v _{P}=\dfrac{d(X_{P}(t))}{dt}$

$v _{P}=a+2bt$

Velocity of Q= $v _{Q}=\dfrac{d(X_{Q}(t))}{dt}$

$v _{Q}=f+2ft$

When the cars have same velocity,

$v_{P}=v_{Q}$

$a+2bt=f+ft$

The time will be

$t = \dfrac{f-a}{2(1+b)}$

Hence, At this time the cars will have same velocity .