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Question :: In a car race , Car A takes time t less than Car B and passes the finishing point with a Velocity V more than the Velocity which Car B passes the point . Assuming that the car start from rest and travel with constant acceleration A1 and A2 2 , Show that V = t (A1 A2 ).

Hey mate here's your answer...

Answer :: 2nd equation or position - time equation -

S = ut +

$\frac{1}{2?} at {2}$

S ---: Displacement

U ---: initial velocity

A --: Acceleration

T --: Time

--

Let time taken by A is T o

Va =

${a}^{1?} to$

Vb =

${a?}^{2?} (t 6 + t)$

Now ,

Va - Vb = V =

$( {a}^{1?} ) - ( {a}^{2?} )t0 - {a}^{2} t$

From Question ::-

Va - Vb = V =

$( {a}^{1?} ) - {a}^{2?} )t0 - {a}^{2} t$

................(1)..

Xa = Xb =

$\frac{1}{2?} {a}^{1?} {t}^{2?}0 = \frac{1}{2 {a}^{2} ?} (t0 + {t}^{2} )$

$\sqrt{ {a}^{1?} } t0 = \sqrt{ {a}^{2?} } (t0 + {t}^{2} )$

$( \sqrt{ {a}^{1?} } - \sqrt{ {a}^{2} })t0 = \sqrt{ {a}^{2} } t$

................(2)

From (1) and (2).

V=

$( {a}^{1?} - {a}^{2} ) \frac{ \sqrt{ {a}^{2}t } }{ \sqrt{ {a}^{1 - {a}^{2?} ?} } ?} - {a}^{2} t$

(

$( \sqrt{ {a}^{1 ?} } + \sqrt{?} {a}^{2} ) \sqrt{ {a}^{2t - {a}^{2t} } }$

V=

=:: V =

$\sqrt{ {a}^{1?} } {a}^{2t?}$

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