Answer:
The shortest distance is 62.7 m
Explanation:
Given;
coefficient of kinetic friction, μk = 0.98
initial velocity, u = 34.7 m/s
Frictional force on the tire;
Fk = -μkN
where;
N is normal reaction = mg
ma = -μkN
ma = -μkmg
a = -μkg
a = - 0.98 x 9.8 = -9.604 m/s²
The shortest distance in which you can stop an automobile by locking the brakes:
Apply equation of motion;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration of the automobile
0 = 34.7² + 2(-9.604)x
0 = 1204.09 - 19.208x
19.208x = 1204.09
x = 1204.09/19.208
x = 62.7 m
Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m
