Question

A point moving with const. acceleration from A to B in a straight line AB has velocities vo and v at A and B respectively. Find its velocity at C which
is at 1/5th of the distance from A.

​

Answer 1

Answer:

Let the distance between A and B=s and as mentioned C is the midpoint of A and B thus distance between A C and

C B=

2

s

From equation of motion we get distance between AB

v

2

=u

2

+2as

s=

2a

v

2

−u

2

----------(1)

Let speed of the vehicle be pm/s at point C

Again from equation of motion we get,

p

2

=u

2

+2a

2

s

as distance between A and C is

2

s

s=

a

p

2

−u

2

-------(2)

Computing 1 and 2 we get

a

p

2

−u

2

=

2a

v

2

−u

2

p

2

=

2

v

2

+u

2

p=

2

v

2

+u

2