Question

Ans 8 please full question

Answer 1

$u = 12m {sec}^{ - 1} \\ t = 6sec \\ v = 0 \\ a = \frac{v - u}{t} = \frac{0 - 12}{6} = - 2m {sec}^{ - 2} \\ s = ut + \frac{1}{2} a{t}^{2} \\ s = 12 \times 6 + \frac{1}{2} ( - 2) {6}^{2} \\ s = 72 - 36 = 36m$

Answer 2

Hi there !!
Here your answer :

Given,

The car is moving with a velocity of 6m/s ,
since it's in motion and is going to come at rest after sometime,

Initial velocity = u = 12m/s
Final velocity = v = 0m/s
Time = t = 6 sec

We know,

i)

$a = \frac{v - u}{t}$

$a = \frac{0 - 12m {s}^{ - 1} }{6s {}^{ - 1} }$

$a = \frac{ - 12}{6}$
a = -2m/s²

Thus,
the acceleration is -2m/s²

ii) To find the distance covered, we will use the following equation of motion

$s = ut + \frac{1}{2} at {}^{2}$
where s is the distance, u is the initial velocity , t is the time and a is the acceleration

$s = 12 \times 6 + \frac{1}{2} \times - 2 \times {6}^{2}$
s = 72 - 36 m

s = 36m

Thus,
the distance is 36 m