Question

A car of mass 1,200 kg is travelling along a straight horizontal road at a speed of 20 m/s when it brakes sharply then skids. Friction brings the car to rest. If the coefficient of friction between the tires and road is 0.8, calculate: a. the deceleration (Answer: -7.84 m/s2)

Answer 1

Answer:

The deceleration of the car is $-7.84\ m/s^2$

Explanation:

It is given that,

Mass of the car, m = 1200 kg

Initial speed of the car, u = 20 m/s

Friction brings the car to rest, final speed, v = 0

The coefficient of friction between the tires and road is 0.8

Let a is the deceleration of car due to force of friction. The frictional force is given by :

$f_k=\mu mg$

$a=-\dfrac{f_k}{m}$

$a=-\dfrac{\mu mg}{m}$

$a=-\mu g$

$a=-0.8\times 9.8$

$a=-7.84\ m/s^2$

So, the deceleration of the car is $-7.84\ m/s^2$. Hence, this is the required solution.

Answer 2

Answer:

- 7.84 m/s^2

Explanation:

m = 1200 kg

u = 20 m/s

v = 0 m/s

μ = 0.8

the deceleration is goiven by

a = - μ x g - 0.8 x 9.8 = - 7.84 m/s^2

Thus, the deceleration is given by - 7.84 m/s^2.