Question

You are driving at the speed of 33.4 m/s (74.7296 mph) when suddenly the car in front of you (previously traveling at the same speed) brakes. Considering an average human reaction, you press your brakes 0.424 s later. Assume that the brakes on both cars are fully engaged and that the coefficient of friction is 0.862 between both cars and the road.
The acceleration of gravity is 9.8 m/s2 .
a) Calculate the acceleration of the car in front of you when it brakes.
b) Calculate the braking distance for the car in front of you.
c) 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 1

Answer:

Part a)

a = - 8.45 m/s/s

Part b)

$d = 66 m$

Part c)

$d = 14.16 m$

Explanation:

Part a)

when car apply brakes then the friction force on the car in front of us is given as

$F_f = \mu mg$

here we need to find deceleration due to friction

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

$a = -\mu g$

$a = -8.45 m/s^2$

Part b)

Braking distance of the car is the distance that it move till it stops

so we will have

$v_f^2 - v_i^2 = 2ad$

$0 - 33.4^2 = 2(-8.45)d$

$d = 66 m$

Part c)

Since we know that average reaction time for human is 0.424 s

now we know that during reaction time our car will travel at uniform speed

so we will have

$d = vt$

$d = (33.4) (0.424)$

$d = 14.16 m$