Question

The initial velocity of a particle moving along a straight line is 5 m/s and its retardation is 1 ms–2. The distance moved by the particle in fourth second of its motion is

Answer 1

Answer:Distance traveled by the body in 4th sec=

U=5m/s , A= -1 m/s^2

Formula for distance traveled in nth sec = u + a{n-1/2}

=5 - 1{4- 1/2}

=3/2

=1.5 m

Therefore the distance traveled by the particle in 4 th sec of its journey is 1.5m.

Hope the answer was helpful.