Question

Brakes applied to car produce an acceleration of 6m/s^2 in opposite direction to motion if the car takes 2s to stop after application of brakes then calculate the distance it travells during this time

Answer 1

Here ,

u(initial velocity) = ?

v(final velocity) = 0m/s

t (Time) = 2 seconds

a (Acceleration) = -6m/s²

So , Use the formula of kinematics to find u.

v = u + at

or, 0 = u + (-6)(2)

or, 0 = u + (-12)

or, 0 = u - 12

or, 12 = u

or, u = 12

Now , we have got Initial velocity , To find distance (s) , we will again apply formula of Kinematics→

s = ut + (1/2)at²

or, s = (12)(2) + (1/2)(-6)(2)²

or, s = 24 + (1/2)(-24)

or, s = 24+(-12)

or, s = 12m

Answer → The distance travelled by car is 12m.
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Answer 2

$\sf \pmb{Answer :}$

12 m

Explanation:

Acceleration produced on applying brakes

= -6 m/(s)^2

Time taken to stop = 2 s

Let initial velocity be u m/s

final velocity is 0 m/s

Using equation of motion

v = u + at

0 = u +(-6)×(2)

u = 12 m/s

Now, using

(v)^2 - (u)^2 = 2as

(0)^2 - (12)^2 = 2 (-6)(s)

-144 = -12(s)

s = 12 m