Question

the brakes applied to a car produce an acceleration of 6m/s in the opposite direction to motion, if the car takes 2s to stop after applying the brakes, calculate the distance it travels during this time.​

Answer 1

$\huge{\mathscr{\fcolorbox{navy}{orange}{\color{darkblue}{ᵃⁿˢʷᵉʳシ︎}}}}$

: 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

Answer 2

Explanation:

Given:-

Acceleration= -6m s-²

Time taken= 2s

Velocity= 0 m s-¹

v = u+ at

0 = u+ (-6m s-²)× 2s

u = 12m s-¹

s= ut+ 1/2 at²

(12m s-¹) × (2s) + 1/2 (-6m s-²) (2s)²

24 m - 12 m

12 m

Therefore , The car will move 12 m before it stops after acceleration.