Question

the breakes applied to a car produces an acceleration of 6 m s^-2 in the opposite direction to the motion .if the car takes 2s to stop after the application of brakes, calculate the distance it travel during this time
.
plz don't spam ​

Answer 1

Acceleration a = -6 m/s^2

Time t = 2 s

Final velocity v = 0 m/s

Let initial velocity be u

Let distance be s

v = u + at

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

So, u = 12 m/s

Now, s = ut + (1/2) at^2

So, s = 12(2) + (1/2)(-6)(2^2)

So, s = 24 - 12

So, s = 12 m

Thus, distance travelled is 12 m

Mark as brainliest ✌️

Answer 2

Answer:

v= u+at

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

or, u= 12m/s^2

now, s= ut + 1/2 at^2

or, s= 12(2) + 1/2(-6) (2) ^2

or, s= 24 -12

or, distance = 12m