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In a car race, car P takes time t(0) les...

In a car race, car P takes time `t_(0)` less than car Q and passes the finishing point with a speed `v_(0)` more than the speed with which xar Q passses the finishing point. Assume that both cars start from rest and trasvel with constant acceeleration `alpha` and `beta` . Find `v_(0)` .

A

`v_(0) = sqrt((alpha^(2)-beta^(2))/(2)t_(0)`

B

`v_(0) = ((alpha+beta)/(2))t_(0)`

C

`v_(0) = sqrt(alpha beta)t_(0)`

D

`v_(0) = sqrt((alpha^(2)+beta^(2))/(2)t_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
In a car race ……….

P.given `t_(Q) - t_(p) = t_(0) - 0 V_(p) - V_(Q) = V_(0)` ………(2)
`x =0 xxt_(p) + (1)/(2)alpha t_(p)^(2) implies t_(p) = sqrt((2x)/(alpha))`
`v_(p)^(2) = 0^(2) + 2alpha x implies V_(p) = sqrt(2alpha x)`
`Q. x = 0 xxt_(Q) + (1)/(2)beta t_(Q)^(2)`
`t_(Q) = sqrt((2x)/(beta))`
now ` (V_(p) - V_(Q))/(t_(Q) - t_(p)) = (v_(0))/(t_(0)) implies (sqrt(2alpha x)-sqrt(2beta x))/(sqrt((2x)/(beta))-sqrt((2x)/(alpha))) = (v_(0))/(t_(0))`
`implies v_(0) = sqrt(alpha beta n t_(0))`
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