Math, asked by janardanbaruah, 8 months ago

An object is pushed along a horizontal surface in such a way that it starts with a velocity of
12 m/s. Its velocity decreases at a rate of 0.5 m/s^2. Find:
a. The time it will take to come to rest.
b. Distance covered by it before coming to rest.​

Answers

Answered by aasarafood24
1

Answer:

Given

v=12m/s

acceleration =0.5m/s^2

u=0m/s

time= v-u/a

time=12-0/0.5=6seconds

diatance= ut 1/2 at^2

distance =0×6×1/2 ×0.5×6^2

distance =0.5×4.5=2.25m

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Answered by kavyasah245
1

a. from 1st equation of motion,

v= u + at (v= 0, u=12m/s, a= -0.5m/s2, t= ?)

0=12+(-0.5* t)

t= 12/0.5

t= 24s

b. from 3rd equation of motion

2as= v2-u2

2*(-0.5)s=0^2 - 12^2

s=24/1

s= 24m

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