Question

The brakes are applied to a car produce an acceleration of 6m/s^2 in the opposite direction to the motion. If the car takes 2 second to stop after the application of brakes, calculate the distance it travels during this time.

Answer 1

A=-6m/s^2 acceleration is negative beacause of yhe it's brakes applied to car .
t=2sec
v=0m/s
u?
s?
v=u+at
u=v-at
u=0-(-6×2)
u=12m/s

s=v^2-u^2/2a
s= 0-144/-12
s=12m

Answer 2

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Note Sign convention is follow in question

$Acceleration \:(a) = 6m\:sec^{-2} (in\: opposite\: direction\:to \: the \: motion\:)\\ a = -6m \: sec^{-2} \\ time \:taken (t) = 2 \: sec\\ Final\: velocity = 0 \: m\: sec^{-1}$

We have v = 0 m/sec

Unknown data

u = ?
s = ?

Formula of motion

$v =\: u\: + \:at$

$s = \: ut + \frac{1}{2} at^{2}$

$2as =\: v^{2} - u^{2}$

Choose one formula

In 2 & 3 formula ( two things are unknown)

[PART - A]

1 is appropriate

$v = u \: + \: at \\ 0 = u + [-6(2)] \\ 0 = u \: -\: 12 \\ u = 12 \:m\: sec^{-1}$

[PART - B]

s = ?

from 2 & 3 choose any 1

I choose ( 2 )

$s = ut +\frac{1}{2} at^{2} \\ s = 12\times 2 +\frac{1}{2} (-6) (2)^{2} \\ s = 24 - 3\times 4 \\ s = 24-12 \\ s = 12 \:m$

$\red{\underline{Answer}}$

Distance travel during retardation when applying breaks becomes 12 m