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In a car race, car A takes 4 s less than...

In a car race, car `A` takes `4 s` less than can `B` at the finish and passes the finishing point with a velocity `v` more than the car `B` . Assuming that the cars start form rest and travel with constant acceleration `a_(1)=4 m s^(-2)` and `a_(2) =1 m s^(-2)` respectively, find the velocity of `v` in m `s^(-1)`.

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The correct Answer is:
8
`t_(1)=t_(2)-t,v_(1)v_(2)=v,S=(1)/(2)a_(1)t_(1)^(2),S=(1)/(2)a_(2)t_(2)^(2)`
`v_(1)=a_(1)t_(1),v_(2)=a_(2)t_(2)impliesv_(2)+v=a_(1)t_(1)`
`impliesa_(2)t_(2)+v=a_(1)t_(1)=a_(1)t_(2)impliest_(2)=(v+a_(1)t)/(a_(1)-a_(2))`
`sqrt((a_(2))/(a_(1)))=(t_(1))/(t_(2))=1-(t)/(t_(2))rArr sqrt((a_(2))/(a_(1)))=1-(t(a_(1)-a_(2)))/((v+a_(1)t))`
`implies(sqrt(a_(2)))/(sqrt(a_(1)))=(v+a_(2)t)/(v+a_(1)t)impliessqrt(a_(2))v+a_(1)sqrt(a_(2)t)=vsqrt(a_(1))+a_(2)sqrt(a_(1)t)`
`impliesv=(sqrt(a_(1)a_(2)))t=8ms^(-1)`
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Knowledge Check

  • In a car race, car A takes a timet 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_1sqrt(a^2))t`
    D
    `(sqrt(a_1a_2))t`

  • In a car race, car A takes t_0 time less to finish than car B and passes the finishing point with a velocity v_0 more than car B . The cars start from rest and travel with constant accelerations a_1 and a_2 . Then the ratio (v_0)/(t_0) is equal to.

    A
    `(a_1^2)/(a_2)`
    B
    `(a_1 + a_2)/(2)`
    C
    `sqrt(a_1 a_2)`
    D
    `(a_2^2)/(a_1)`

  • In a car race car A takes t_(0) time less to finish than car B and pases the finishing point with a velocity v_(0) more than car B . The cars start from rest and travel with constant accelerations a_(1) and a_(2) . Then the ratio (v_(0))/(t_(0)) is equal to

    A
    `(a_(1)^(2))/(a_(2))`
    B
    `(a_(1)+a_(2))/2`
    C
    `sqrt(a_(1)a_(2))`
    D
    `(a_(2)^(2))/(a_(1))`