Physics, 04.07.2019 09:30 bryce99

Homework (part 1 of 3) you are driving at the speed of 25.6 m/s (57.6035 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 0.731 s later. you also brake and decelerate as rapidly as possible without skidding. assume that the coefficient of static friction is 0.981 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. (part 2 of 3) calculate the braking distance for the car in front of you. answer in units of m. (part 3 of 3) find the minimum safe distance at which you can follow the car in front of you and avoid hitting it (in the case of emergency braking described here). answer in units of m.

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Answer from: tia5520

This question involves the concepts of the equations of motion, frictional force, Newton's Second Law and uniform speed.

(1) The acceleration of the car "- 9.62 m/s²".

(2) The braking distance of the car in front is "34.06 m".

(3) The minimum safe distance between the cars is "18.71 m".

(1)

The acceleration (deceleration) of the car can be calculated using the formula for frictional force:

$F = -\mu mg$ (negative sign due to opposite direction to motion)

using Newton's Second Law:

$ma = -\mu mg\\a = -\mu g$

where,

a = acceleration = ?

μ = coefficcient of friction = 0.981

g = acceleation due to gravity = 9.81 m/s²

Therefore,

$a = -(0.981)(9.81\ m/s^2)\\$

a = -9.62 m/s²

(2)

The braking distance can be found using the third equation of motion:

2as = v_f^2-v_i^2

where,

s = braking distance = ?

vf = final speed = 0 m/s

vi = initial speed = 25.6 m/s

Therefore,

$2(-9.62\ m/s^2)s=(0\ m/s)^2-(25.6\ m/s)^2\\\\s = \frac{-655.36\ m^2/s^2}{-19.24\ m/s^2}\\\\$

s = 34.06 m

(3)

The safe distance will be equal to the extra distance travelled by car before applying the brake, during the reaction time. Since it is the uniform speed motion, hence using the equation:

$d = vt\\d = (25.6\ m/s)(0.731\ s)$

d = 18.71 m

Learn more about equations of motion here:

link

The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer from: hHRS8801

Part 1 of 3

Force of friction due to applied brakes is given as

$F_f = \mu mg$

now we will have acceleration given by equation

$a = \frac{F_f}{m}$

$a = \frac{\mu m g}{m}$

$a = \mu g$

now plug in the above values

a = 0.981* 9.8 = 9.61 m/s^2

Part 2 of 3

now we can find the braking distance by using kinematics

v_f^2 - v_i^2 = 2 a d

here we know that

v_f = 0

v_i = 25.6 m/s

a = -9.61 m/s^2

now by above equation

0 - 25.6^2 = 2(-9.61)d

d = 34.1 m

Part 3 of 3

now the car behind the first car will move with constant speed for the reaction time

d = v*t

d = 25.6*0.731 = 18.7 m

so minimum safe distance between the two cars must be equal to above distance

d = 18.7 m

Homework (part 1 of 3) you are driving at the speed of 25.6 m/s (57.6035 mph) when suddenly the car

Answer from: Quest

ill do it for 100, also this isnt 80 its 40

Answer from: Quest

graph question answered .. you think better when you are relaxed and happy

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Homework (part 1 of 3) you are driving at the speed of 25.6 m/s (57.6035 mph) when suddenly the car...

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