JEE Physics/ Lakshmikanta Satapathy/ Kinematics QA part 8/ Question on uniformly accelerated motion and average velocity solved with the related concepts

1. Physics Helpline
L K Satapathy
Kinematics QA 8
Uniformly Accelerated Motion
BCA
S S
2S
u x v

2. Physics Helpline
L K Satapathy
Kinematics QA 8
Answer :
Question : A particle , moving with uniform acceleration from A to B along a
straight line , has velocities u and v at A and B respectively. If C be the mid point
of AB and the time taken by the particle from A to C is twice that from C to B,
then
(a) v = 3u (b) 3v = 7u (c) 2v = 7u (d) v = 7u
Given : Velocity at A = u
& Velocity at B = v
Let the distances AC = CB = S  AB = 2S
Let the uniform acceleration = a
BCA
S S
2S
u x vThe situation is shown in the figure
Let velocity at C = x
Given : Time taken for AC = 2  Time taken for CB

3. Physics Helpline
L K Satapathy
Kinematics QA 8
For AC : Initial velocity = u , Final velocity = x & time = 2t
2 . . . (1)x u at  
For CB : Initial velocity = x , Final velocity = v & time = t
. . . (2)v x at  
(1)&(2) 2( ) 2 2x u v x v x     
3 2x v u  
2 . . . (3)
3
v ux  
Let the time taken for CB = t  Time taken for AC = 2t
 Time taken for AB = 3t

4. Physics Helpline
L K Satapathy
Kinematics QA 8
Correct option = (d)
3 6 2v u v u   
For uniformly accelerated motion , distance covered = average velocity  time
 For AC : 2 ( ) . . . (4)
2
u x SS t u x
t
    
& For CB : . . . (5)
2 2
x v S x vS t
t
    
(4)&(5) ( )
2
x vu x   
2 2u x x v   
2 . . . (6)x v u  
2(3) &(6) 2
3
v u v u  
[7 ]sv u An 

5. Physics Helpline
L K Satapathy
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