Question

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

Answer 1

Answer:

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

Answer 2

Answer:

Explanation:

Solution,

Final velocity, v = 0 (As brakes applied)

Acceleration, a = 6 m/s²

Time taken, t = 2 seconds

To Calculate,

Distance traveled, s = ??

According to the 1st equation of motion,

We know that

v = u + at

So, putting all the values, we get

⇒ v = u + at

⇒ 0 = u + 6 × 2

⇒ u = 6 × 2

⇒ u = 12 m/s

Here, the initial velocity is 12 m/s.

Now, we will find distance traveled,

According to the 3rd equation of motion,

We know that,

v² - u² = 2as

So, putting all the values again, we get

⇒ v² - u² = 2as

⇒ (0)² - (12)² = 2 × 6 × s

⇒ 144 = 12s

⇒ 144/12 = s

⇒ s = 12 m.

Hence, the distance traveled by car is 12 m.