Question

A particle 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/s2

. (a) Find the time it will take to come to rest. (b)

Find the distance covered by it before coming to rest.​

Answer 1

Answer:

your answer is ready sachin phogat

Explanation:

initial velocity of particle  = 12 m/sec

negative acceleration  = -0.5 m/sec ² [ negative acceleration show negative sign ]

final velocity                  = 0 m/sec   [ particle finally come to rest ]

now, by defination of acceleration

a = v - u / t

-0.5  = 0 - 12 /t

t     = -12/-0.5   =  24 sec

then by second equation of motion

s = ut + 1/2 at²

s = 12×24 + 1/2×-0.5×(24)²

s = 288 + 1/2×-0.5×576

s = 288 +288×0.5

s = 288 + 144 = 432m

you can also find displacement by third equation of motion that is

2as = v² - u²

hope it helps you very well

sachin mark my answer as brainlist

I am mohit dangi