Question

A body travels along a straight path ABC. At A its velocity is 5 m/s and at C its velocity is 15 m/s and AB:BC is 3:5. If it takes 15 sec to travel AB, What will be the time taken to travel BC

Answer 1

Answer:

The velocity at C is 15 m/s.

Explanation:

Given that,

Initial velocity  = 6 m/s

Final velocity  = 9 m/s

The ratio of AB:BC=5:16

A particle is moving with uniform acceleration along a straight line ABC its velocity at A and B are 6 m/sec and 9 m/sec respectively.

Let be the distance AB= 5 x and BC = 16 x

Then,  for A to B

Using equation of motion

v^2-u^2=2asv2−u2=2as

9^2-6^2=2\times a\times5x92−62=2×a×5x

ax = 4.5ax=4.5 ....(I)

The velocity at C is

For B to C

v^2-9^2=2\times a\times16xv2−92=2×a×16x

v^2-81=32axv2−81=32ax ....(II)

Put the value of ax in equation (II)

v^2=32\times4.5+81v2=32×4.5+81

v=\sqrt{225}v=225

v = 15\ m/sv=15 m/s

Hence, The velocity at C is 15 m/s