Question

A car is moving with a I 16 m/s
when breake is applied . The force
applied by the breaker is
1000 N. The mass of the car
to with is passangers in 1200 kg a. how long should the brakes be applied to make the car come to a hault. b. how far does the car travel before it comes to rest.​

Answer 1

Deceleration provided by brake = 1000/1200 = 1/12 m/$s^{2}$

V=0 , u = 16m/s, a = -1/12

v = u+at => 0 = 16 - 1/12 x t

t = 16*12 seconds = 192 SECONDS brake should be applied.

for how far car travel use any

s=ut + 1/2 a$t^{2}$

or

$v^{2}  = u^{2}  + 2 aS$

Answer 2

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

from the newtons laws of motion

f=ma

now...f=1000N

m=1200kg

therefore...a=1000/1200=(5/6)m/s²

now....final velocity is zero....

so....

:=u -at

0=16-(5/6)t

t=96/5

t=19.2sec (almost)

now....

s=ut-0.5at²=16(19.2)-0.5(5/6)19.2²

=>s=307.2-153.6=153.6 m