IIT JEE , AIEEE Forum - Physics , Mechanics

kaka01

A particle starts from rest and traverses a distance s with uniform acceleration and then moves uniformly with the acquired velocity over a further distance 2s. Finally it comes to rest after moving through a further distance 3s under uniform retardation. Assuming the entire path is a straight line, then the ratio of the average speed over the journey to the maximum speed on way is ?

a) 5/3 b) 2/5 c 3/5 d ) 5/2

ashish17

How can a body comes to rest twice????????

please check the ques.

kaka01

My mistake sorry , i have corrected the question , now try to solve .

kaka01

Ok if you dont want to give me sol thats ok, give me answer at least.

sahilmadaan12

ans is c 3/5
wait m typing the solution

sahilmadaan12

as u can see that whatever will be the speed attained after !st acceleration will be the maximum speed......
So let the maximum speed be v
now average speed = (total distance)/total time
total distance = s + 2s + 3s= 6s
now to find the total time separate it in phases
now in first phase
v^2 = 2as let a be the acceleration
v= a*T1 T1be the time in first phase
thus
T1=2s/v

phase 2
constant speed
time T2= 2s/v

phase 3

0=v^2 + 2*b*3s
0=v+b*T3
T3=6s/v

therefore T1 + T2 +T3 = 10S/V
THEREFORE AVERAGE SPEED = 6V/10 =3V/5
AND HENCE WE GET THE RATIO AS 3/5

kaka01

Thanks buddy , it was a great help . thanks a million times .

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