Question

A particle moving with constant acceleration of 2m/s² due west has an initial velocity of 9m/s due east . Find the distance covered in the fifth second of its motion

Answer 1

Answer:

Here the particle velocity will become zero in 4.5 seconds, as

v=u+at

0=9−2×t

t=4.5sec

Displacement in the same period will be

v

2

−u

2

=2as

−81=2s

s

4.5

=20.25m

Let us call the point of velocity being zero as Point A.

now body will start moving backwards,

displacement in next 0.5 sec from point A,

s=ut+

2

1

at

2

s

0.5

=

2

1

(−2)(0.25)

2

=−0.25m

hence distance covered is 0.25 m

(-ve sign indicates body moved in backward direction. It is to be noted that displacement and distance are not to be confused to be same)

Distance traveled in 4 seconds

s

4

=9×4+

2

1

(−2)(4)

2

s

4

=36−16=20

Thus,

distance covered in 5th second is

D=(s

4.5

+∣s

0.5

∣−s

4

)=20.5+0.25−20=0.5