A particle starts from rest and travel a distance S with a uniform acceleration and them moves uniformly with the aquired velocity over a further distance 2 S. Finally it comes to rest after moving through a further distance 3 S under uniform deacceleration. 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?

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2015-08-22T20:50:58+05:30

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Let a be the uniform acceleration.

  v² - u² = 2 a S    =>  S = v² / 2 a        as u = 0.
  final velocity =  v = √(2 a S)
  time taken to reach this velocity :  t1 = (v - u ) / a  = v / a = √(2S/a)

Now the particle moves with this velocity for distance 2 S.
   so time taken t2 = 2 S / v = 2 S / √(2aS) =  √(2 S / a)

Now the particle moves with a uniform deceleration  - a3 for a distance 3 S.
     time taken t3 to stop.
           2 * a3 * 3 S = v² - u²         here v = 0  and u = √(2 a S)
           =>  a3 =  a / 3 
            =>  t3 = - u /a3 =  3 √( 2 S / a)

total time duration =  t = t1 + t2 + t3 = 5 * √(2 S / a)
total distance travelled :  S + 2S + 3S =6 S

   => average speed = 3 / 5 * √(2 a S)
   maximum speed =  √(2 a S)

Ratio =  3/5
 

2 5 2
what is the formular for average speed ?
v-u/t
ok i got it
avg speed formula = total distance travelled / total time duration..
I thought acceleration