Question

12. In a car race, car A takes 2 s less than car B and
passes the finishing point with a velocity v more
than the velocity with which car B passes the point.
Both the cars start from rest and travel with constant
accelerations of aA = 3 m/s² and aB = 2 m/s². The value of v is _
​

Answer 1

Answer:

Explanation:

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 (v1  +v). Writing equations of motion for A and B.

v  1 +v=a 1  t  1               equation 1  

v  1  =a  2  (t  2 +t)          equation 2

From equations (1) and (2 ), we get

v=(a  1  −a  2  )t  1  −a  2 t          equation 3

Total distance travelled by both the cars is equal

S  A  =S  B

⇒  1/2  a  1  t  1  ^(2)  =  1/2  a  2  (t  1 +t)  ^(2)

⇒t  1 = [root(a2) x t] / [root (a1) - root (a2)]

Substituting this value of t  1  in equation (3), we get

v=[root (  a  1  a  2  )] x t

By this formula,

v = [root ( 3 x 2 ) ] x 2

v= 2 x (root 6 )

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Answer 2

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