Question

In a car race, car A takes a time T sec less than the car B and passes the finishing part with a velocity v more than that of the car B. If the cars start from rest and travel with constant accleration $A_1$ and $A_2$ respectively. Show that $v=T\sqrt{A_1{A}_2}$.

Answer 1

Car	Time	Velocity	Acceleration
A	T1	v1	A1
B	T2	v2	A2

It is given that :

$T_1+t=T_2$

$v_1=v_2+v$

Now using equation of motion :

$s_1=u_1{T}_1+\frac{1}{2}{A}_1{T}_1^2$

$\Rightarrow s=\frac{1}{2}{A}_1{T}_1^2$ ...(1)

$s_2=u_2{T}_2+\frac{1}{2}{A}_2{T}_2^2$

$\Rightarrow s=\frac{1}{2}{A}_2{T}_2^2$ ...(2)

From equation (1) and (2)

$s_1=s_2$

$\frac{1}{2}{A}_1{T}_1^2=\frac{1}{2}{A}_2{T}_2^2$

$\Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{A_1}{A_2}}$

$\Rightarrow \frac{T_1+T}{T_1}=\sqrt{\frac{A_1}{A_2}}$

$\Rightarrow 1+\frac{T}{T_1}=\sqrt{\frac{A_1}{A_2}}$...(3)

$v_1=u_1+A_1{T}_1=A_1{T}_1$ ...(4)

$v_2=u_2+A_2{T}_2=A_2{T}_2$.......(5)

$v_1^2=u_1^2+2A_{1}s=2A_{1}s$

$v_2^2=u_2^2+2A_{2}s=2A_{2}s$

$\frac{v_1^2}{v_2^2}=\frac{2A_{1}s}{2A_{2}s}$

$\Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{A_1}{A_2}}$

$\Rightarrow \frac{v_2+v}{v_2}=\sqrt{\frac{A_1}{A_2}}$

$\Rightarrow 1+\frac{v}{v_2}=\sqrt{\frac{A_1}{A_2}}$ ...(6)

From equation (3) and (6), we get

$\frac{v}{v_2}=\frac{T}{T_1}$

$\Rightarrow \frac{v}{A_2{T}_2}=\frac{T}{T_1}$ (from eq.(5))

$\Rightarrow v=\frac{T{A}_2{T}_2}{T_1}=T{A}_2\sqrt{\frac{A_1}{A_2}}$

$\Rightarrow v=T\sqrt{A_1{A}_2}$.