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Question

Physics Question on speed and velocity

Two cars PP and QQ start from a point at the same time in a straight line and their positions are represented by xP(t)=at+bt2x _{ P }( t )= at + bt ^{2} and xQ(t)=ftt2x _{ Q }( t )=f t - t ^{2}. At what time do the cars have the same velocity ?

A

af1+b\frac{a - f}{1 + b}

B

a+f2(b1)\frac{a + f}{2( b - 1)}

C

a+f2(1+b)\frac{a + f}{2( 1 + b)}

D

fa2(1+b)\frac{f - a }{2( 1 + b)}

Answer

fa2(1+b)\frac{f - a }{2( 1 + b)}

Explanation

Solution

vp=dxpdt=a+2btv_{p} = \frac{dx_{p}}{dt} = a +2bt vQ=dxQdt=f2tv_{Q} = \frac{dx_{Q}}{dt} = f - 2t vP=vQa+2bt=f2t v_{P} = v_{Q} \Rightarrow a +2bt =f-2t t=fa2(b+1)t =\frac{ f-a}{2\left(b+1\right)}