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 $