Question

In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of the car B. Assuming that both the cars starts from rest and travel with constant acceleration $a_1$ and $a_2$ respectively. Show that $v=\sqrt{a_1{a}_2}t$

Answer 1

Let the distance between the starting and the finishing point be $x$.

$\therefore$ $t_1=t_2-t$ and $V_1=V_2+v$ ........(a)

Car A :

$V_1=0+a_1{t}_1$ $⟹V_1=a_1{t}_1$ .........(1)

Also $x=0+\frac{1}{2}{a}_1{t}_1^2$

$⟹x=\frac{a_1{t}_1^2}{2}$ ........(2)

Car B :

$V_2=0+a_2{t}_2=a_2(t+t_1)$ .........(3)

Also $x=0+\frac{1}{2}{a}_2{t}_2^2$ $⟹$ $x=\frac{a_2{t}_2^2}{2}$ ..........(4)

From (1), (3) & (a) we get $a_1{t}_1=a_2(t+t_1)+v$ $⟹t_1=\frac{a_{2}t+v}{a_1-a_2}$

Equating (2) and (4), $a_1{t}_1^2=a_2{t}_2^2$

$\therefore$ $a_1{t}_1^2=a_2(t+t_1{)}^2$ $⟹(a_1-a_2)t_1^2-2a_{2}t{t}_1-a_2{t}^2=0$

OR $(a_1-a_2)\times \frac{(a_{2}t+v{)}^2}{(a_1-a_2{)}^2}-2a_{2}t\times \frac{a_{2}t+v}{a_1-a_2}-a_2{t}^2=0$

OR $(a_{2}t+v{)}^2-2a_{2}t(a_{2}t+v)-a_2(a_1-a_2)t^2=0$

OR $a_2^2{t}^2+v^2+2a_{2}vt-2a_2^2{t}^2-2a_{2}vt-a_1{a}_2{t}^2+a_2^2{t}^2=0$ $⟹v=\sqrt{a_1{a}_2}$ $t$