contestada

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same speed) brakes and begins to slow down with the largest deceleration possible without skid- ding. Considering an average human reaction, you press your brakes 0.507 s later. You also brake and decelerate as rapidly as possible without skidding. Assume that the coefficient of static friction is 0.868 between both cars’ wheels and the road. The acceleration of gravity is 9.8 m/s2 . Calculate the acceleration of the car in front of you when it brakes. Answer in units of m/s2.

Respuesta :

Here when car in front of us applied brakes then it is slowing down due to frictional force on it

So here we can say that friction force on the car front of our car is given as

[tex]F_f = \mu m g[/tex]

So the acceleration of car due to friction is given as

[tex]F_{net} = - \mu mg[/tex]

[tex]a = \frac{F_{net}}{m}[/tex]

[tex]a = -\mu g[/tex]

now it is given that

[tex]\mu = 0.868[/tex]

[tex]g = 9.81 m/s^2[/tex]

so here we have

[tex]a = -0.868 * 9.81[/tex]

[tex]a = -8.52 m/s^2[/tex]

so the car will accelerate due to brakes by a = - 8.52 m/s^2

The acceleration of the car when it was front of him is [tex]\\\rm \bold { -8.52 m/s^2}[/tex].

Given here,

[tex]\rm \bold{ \mu = 0.868}\\\rm \bold{ g =9.81 m/s^2}[/tex]

The friction force applied on car is

[tex]\rm \bold { F_f= \mu mg}\\[/tex]

[tex]\rm \bold { F_n_e_t= - \mu mg}\\[/tex]

The acceleration due to friction

[tex]\rm \bold { a = \frac{F_n_e_t}{m} }\\\\\rm \bold { a = -\mu g}[/tex]

[tex]\rm \bold { a = -0.868\times 9.81}\\\rm \bold { a = -8.52 m/s^2}[/tex]

Hence, we can conclude that the acceleration of the car when it was front of him is [tex]\\\rm \bold { -8.52 m/s^2}[/tex].

To know more about acceleration, refer tot the link:

https://brainly.com/question/12550364?referrer=searchResults

Otras preguntas