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

Two cars P and Q start from a point at the same time in a straight line and their position are represented by x_p(t) = at + bt^2 and x_Q (t) = ft - t^2 . At what time do the cars have the same velocity ?

Two cars A and B travel in a straight line.The distance of A, from the starting point is given as a function of time by x_(A)(t)=at+bt^(2) with a=4m/s and b=2m/s^(2) .The distance of B from the starting point is given as x_(B)(t)=ct^(2)+dt^(3) with c=2m/s^(2) and d=1m/s^(3) . Which car moves ahead just after they leave the starting point?

The position of a particle moving on a straight line depends on time t as x=(t+3)sin (2t)

A p article moves in a straight line such that its position in 't' time is s(t)=(t^(2)+3)/(t-1) cm. find its velocity at t=4 sec.

The diagram shows a velocity-time graph for a car moving in a straight line. At point Q the car must be

The position of a point in time t is given by x=a+bt-ct^(2),y=at+bt^(2) . Its acceleration at time t is

The position of a point in time t is given by x=a+bt-ct^(2),y=at+bt^(2) . Its acceleration at time t is

Initially car A is 10.5 m ahead of car B . Both start moving at time t = 0 in the same direction along a straight line. The velocity-time graph of two cars is shown in figure. The time when the car B will catch the car A , will be.

A point moves in the straight line during the time t=0 to t=3 according to the law s=15t-2t^2 . The average velocity is