Question

A particle moving with a velocity of 20 m/s is decelerated uniformly at the

rate of 1 m/s2

Find (i) The distance covered by the particle to the end of 10 second

(ii) The time taken by it to come to rest . Please give answer with explanation. ​

Answer 1

Answer:

10m/s will be the velocity

20 sec will be required to put it at rest

Answer 2

Correct question:-

A particle moving with a velocity of 20m/s is decelerated uniformly at the rate of 1m/s².Find:-

(i)The distance covered by the particle at the end of 10s

(ii)Time taken by it to come to rest.

Given:-

→Initial velocity of the particle=20m/s

→Deceleration/retardation=1m/s²

→Time=10s

To find:-

→Distance travelled by the particle in 10s

→Time taken by it to come to rest

Solution:-

Here as deceleration of the particle is given as 1m/s²,thus it's acceleration will be -ve,i.e. -1m/s²

By using the 2nd equation of motion,we get:-

=>s=ut+1/2at²

=>s=20×10+1/2×(-1)×(10)²

=>s=200-50

=>s=150m

In this case,to find the time taken by the particle to come to rest,we shall take final velocity of the car as 0 i.e.v=0 as the particle is finally stopping and coming to rest.

Now,by using the 1st equation of motion,we get:-

=>v=u+at

=>0=20+(-1)t

=>-20= -t

=>t = 20s

Thus:-

→Distance travelled by the particle at the end 10s is 150m.

→Time taken by it to come to rest is 20s.