3) Problem Name: Stopping Distance A 1500 kg car is travelling uphill at 30.0 m/

Question

3) Problem Name: Stopping Distance A 1500 kg car is travelling uphill at 30.0 m/s when the driver sees something in the road ahead and slams on the brakes. The car skids and eventually comes to a halt some distance farther along. The coefficient of kinetic friction between the road and the tires is k = 0.800. Assume the acceleration is constant during the entire incident. a) Determine a symbolic expression for the stopping distance d, and the acceleration a, if the car is traveling up an incline at with respect to level ground. b) Find numerical values for d and a for +10.00 slope (up hill) c) Find numerical values for d and a for 6-0 (level ground) d) Find numerical values for d and a for =-10.0° slope (down hill) e) Find numerical values for d and a for = +10.0° frictionless slope (up an icy hill) Make dlear diagram(6, and write clear a list of "knowns: UNDERSTAND)

Explanation / Answer

given,

mass of car = 1500 kg

initial speed of car = 30 m/s

coefficient of kinetic friction uk = 0.8

car is travelling uphill so,

equation of motion will be

mg * sin(theta) + friction force = ma

friction force = uk * mg * cos(theta)

so,

mg * sin(theta) + uk * mg * cos(theta) = ma  

g * sin(theta) + uk * g * cos(theta) = a  

by third equation of motion

v^2 = u^2 + 2ad

since final velocity = 0 m/s

so,

0 = 30^2 - 2 * g * sin(theta) + uk * g * cos(theta) * d

d = 900 / (2 * g * sin(theta) + uk * g * cos(theta))

if theta = 10 degree

a = g * sin(theta) + uk * g * cos(theta)

a = 9.8 * sin(10) + 0.8 * 9.8 * cos(10)

a = 9.422 m/s^2

d = 900 / (2 * g * sin(theta) + uk * g * cos(theta))

d = 900 / (2 * 9.8 * sin(10) + 0.8 * 9.8 * cos(10))

d = 80.903 m

if theta = 0

a = 9.8 * sin(0) + 0.8 * 9.8 * cos(0)

a = 7.84 m/s^2

d = 900 / (2 * 9.8 * sin(0) + 0.8 * 9.8 * cos(0))

d = 114.796 m

if theta = -10 degree

a = 9.8 * sin(-10) + 0.8 * 9.8 * cos(-10)

a = 6.019 m/s^2

d = 900 / (2 * 9.8 * sin(-10) + 0.8 * 9.8 * cos(-10))

d = 208.459 m

if theta = 10 degree and uk = 0

a = 9.8 * sin(10) + 0 * 9.8 * cos(10)

a = 1.7 m/s^2

d = 900 / (2 * 9.8 * sin(10) + 0 * 9.8 * cos(10))

d = 264.43 m

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