Question

A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12m/s. Its velocity decreases at a uniform rate of 0.5m/s2. (a) Find the time it will take to come to rest. (b) Find the distance covered by it before coming to rest (c) Calculate the force acting on it when the mass of it is 10 kg.





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Answer 1

Answer:

a=24seconds

b=144metres

c=-5N

Explanation:

a)initial velocity=12m/s

acceleration=-0.5m/s

final velocity=0m/s

According to 1st eq. of motion

v=u + at

0=12+(-0.5)t

-12/-0.5=t

-120/5=t

24 seconds

b)according to 2nd eq. of motion-

s=ut+at^2/2

=12(24)+(-0.5)(24*24)/2

=288+(-0.5)(288)\

=288-144

=144metres

c)force=mass *acceleration

=10*(-0.5)

=-5N

Answer 2

Answer:

a)24seconds

b)144metres

c)-5N

Explanation: