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A particle moving with uniform acceleration along a straight line passes through successive points P, Q, R where PQ:QR = 3:5 and time taken from P to Q is 40 seconds. If the velocity of the particle at P and R are 10m/s and 30m/s . Find: The velocity of the particle at Q The acceleration of the particle. Time taken to cover Q to R. Total time from P to R.

A particle moving with uniform acceleration along a straight line passes through successive points P, Q, R where PQ:QR = 3:5 and time taken from P to Q is 40 seconds. If the velocity of the particle at P and R are 10m/s and 30m/s. Find:
  1.  The velocity of the particle at Q
  2. The acceleration of the particle.
  3. Time taken to cover Q to R.
  4. Total time from P to R.

Grade:11

2 Answers

Naveen Kumar
36 Points
6 years ago
______________________________
P.                 Q.                             R
←.  3.    →.    ←.       5.           →
 
From P to R
u = 10m/s.    v = 30 m/s.     S =8 m ( let 3 + 5 )
v2 = u2 + 2as
900 = 100 + 16a  ( after putting values)
=> 16a = 800
=> a = 800/16
=> a = 50 m/s2. .            ACCELERATION
 
Therefore velocity v2 at Q is
u = 10m/s.   a = 50m/s2.      S = 3
 
v2 = u2 + 2as
v2 = 100 + 2*50*3
v = 20m/s.                      VELOCITY
 
Time from Q to R
u = 20m/s.           v= 30m/s.      a = 50m/s2
v = u + at
t = v-u/a
t = ( 30 - 20)/50
t= 0.2 seconds.                          TIME
 
Vikas TU
14149 Points
6 years ago
Dear Student,
U might like this one better:

Let the velocity at point Q is x.
PQ:QR = 3:5
let it be 3y and 5y

v2-u2=2as
=>900-x2=2xax5y
=>10ay=900-x2 ------1

velocity of the particle at P and R are 10m/s and 30m/s
solving this by putting v2-u2=2as
gives
900-100=2xax8y
=>ay=50

substituting ay in 1
=>10x50=900-x2
=>x=20m/s
 
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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