Question

A particle has an initial velocity of 18 metres per second and involves a constant deceleration of 4ms^-2. What is the distance travelled in the 5th second?

Answer 1

Answer:

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

v=u+at0=9−2×tt=4.5sec

Displacement in the same period will be

v2−u2=2as−81=2ss4.5=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+21at2s0.5=21(−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

s4=9×4+21(−2)(4)2s4=36−16=20

Thus,

distance covered in 5th second is

D=(s4.5+∣s0.5∣−s4)=20.5+0.25−20=0.5