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.
Answers
Answer:
10m/s will be the velocity
20 sec will be required to put it at rest
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.