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 same velocity

Answer 1

Hie!!

For car p

Xp(t) = at + bt²

Differentiate both sides w.r.t. t we have.

dxp(t)/dt = a + 2bt

Vp = a + 2bt

For car q

Xq(t) = ft - t²

dxq(t)/dt = f - 2t

Vq = f - 2t

According to the given Question!!

f - 2t = a + 2bt

( f - a ) = 2bt + 2t

t = (f - a)/(2b + 1)

Answer 2

Questions:

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 same velocity

Answer:

option d