IIT JEE , AIEEE Forum - Physics , Mechanics

ishan.maheshwari

explain the method to solve the question given below. do not state only the answer.

a car starts from rest and acquires a velocity v with uniform accleration a. then it comes to a stop with uniform retardation b . the average velocity of the car is :

(a) v (b) v / 2 (c) ab / a+b (d) a+b / ab

kusum

let the car take a time t1 to acwuire the velocity v

v=at1

then let it take a time t2 to come to rest

0-v=(-b)t2

the total distance travlled in the trip

1/2at1^2 + vt1 +vt2 - 1/2bt2^2 =s

now the average velocity=total distace / total time
=s/t
here t=t1+t2 .

ishan.maheshwari

but how to get the final answer among the 4 options that i have specified ?

Prakriteesh

v = at₁

i.e. t₁ = v/a

0 = v - bt₂

i.e. t₂ = v/b

So, total time of travel is t₁+ t₂ = v(1/a +1/b)

Now,

v² = 2as₁ or s₁ = v²/2a

0 = v² - 2bs₂ or s₂ = v₂/2b

So, total distance = s₁+ s₂ = v²/2 . (1/a + 1/b)

So, avg. speed = total distance / total time

= v/2

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