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. $(\sqrt{a_1/a_2})t$
• B. $(\sqrt{a_2/a_1})t$
• C. $(a_1\sqrt{a^2})t$
• D. $(\sqrt{a_1{a}_2})t$

Answer 1

Answer: (D)

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) $v_1=a_2(t_1+t)$ ? (ii) From Eqs.(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 Eq. (iii), we get $v=(\sqrt{a_1{a}_2})t$