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 accelerations $a_1$ and $a_2$ respectively. So, the value of $v$ will be :

• A. $\left(\sqrt{a_1/a_2}\right)t$
• B. $\left(\sqrt{a_2/a_1}\right)t$
• C. $\left(a_1\sqrt{a_2}\right)t$
• D. $\left(\sqrt{a_1{a}_2}\right)t$

Answer 1

The correct option is D $\left(\sqrt{a_1{a}_2}\right)t$

Consider that $A$ takes $t_1$ second, then according to the given problem, $B$ will take $(t_1+t)$ seconds. Further let $v_1$ be the velocity of $B$ at finishing point, then velocity of $A$ will be $(v_1+v)$. Writing equations of motion for $A$ and $B$
$v_1+v=a_1{t}_1$ ....(i)
$v_1=a_2(t_2+t)$ ....(ii)
From equations (i) and (ii), we get
$v=(a_1-a_2)t_1-a_{2}t$ .....(iii)
Total distance travelled by both the cars is equal
$S_A=S_B$
$\Rightarrow \frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}{a}_2{(t_1+t)}^2\Rightarrow t_1=\frac{\sqrt{a_2}t}{\sqrt{a_1}-\sqrt{a_2}}$
Substituting this value of $t_1$ in equation (iii), we get
$v=\left(\sqrt{a_1{a}_2}\right)t$