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In a car race straight road, car A takes a time t less than car B at finish and passes finshing point with a speed 'v' more than that of a car B. Both the cars start from rest and travel with the constant acceleration a1 and a2 repectively . Then 'v' is equal to:

 

  • Option 1)

    \sqrt{a_{1}a_{2}t}

  • Option 2)

    \frac{a_{1}+a_{2}}{2}t

  • Option 3)

    \sqrt{2a_{1}a_{2}}t

  • Option 4)

    \frac{2a_{1}a_{2}}{a_{1}+a_{2}}t

Answers (1)

best_answer

 

2nd equation or Position- time equation -

s= ut +\frac{1}{2}at^{2}

s\rightarrow Displacement

u\rightarrowInitial velocity

a\rightarrowacceleration

t\rightarrow time

 

-

Let time taken by A is t_{o}

V_{A}-V_{B}=V=(a_{1}-a_{2})t_{o}-a_{2}t

from question

V_{A}-V_{B}=V=(a_{1}-a_{2})t_{o}-a_{2}t.....................(1)

X_{A}=X_{B}=\frac{1}{2}a_{1}t_{o}^{2}=\frac{1}{2}a_{2}(t_{o}+t)^{2}

\sqrt {a_{1}t_{o}}=\sqrt a_{2}(t_{o}+t)

(\sqrt a_{1} - \sqrt a_{2})t_{o}=\sqrt a_{2} \: t........................................(2)

from (1) and (2)

V=( a_{1} - a_{2})\frac{\sqrt a_{2}t}{\sqrt a_{1}- \sqrt a_{2}}-a_{2}t

(\sqrt a_{1}+\sqrt a_{2})\sqrt a_{2} t -a_{2}t

=>V=\sqrt {a_{1}a_{2}}\: t


Option 1)

\sqrt{a_{1}a_{2}t}

Option 2)

\frac{a_{1}+a_{2}}{2}t

Option 3)

\sqrt{2a_{1}a_{2}}t

Option 4)

\frac{2a_{1}a_{2}}{a_{1}+a_{2}}t

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