Physics, asked by sandy33, 1 year ago

in a car race, car A takes 2s 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 ms^-2 and aB =2 ms^-2. The value of v is

sandy33: m/s now solve

sandy33: ans of this question is 4.9 m/s

Dakshansh: from where did you got it

sandy33: hv u solved it

sandy33: my friend gave it to me

Dakshansh: i think the answer is wrong bcoz,

sandy33: no its the ans

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Answers

Answered by Dakshansh

the answer is wrong bcoz,
a= v-u/t
and u= 0
so a= v/t
it is given that the answer is 4.9
so, t = v/a = 4.9 / 3 = 1.63 s
now the time taken by car b is two more than car a so,
t(of car b)= 1.63 + 2 = 3.63s
so v of car B = a x t = 2 x 3.63 = 7.26 m/s
but it is given that v(of car A) > v(of car B)
so according to me the answer 4.9 is wrong

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