Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.
Answer:
2.99 m/s
Explanation:
Stopping distance, s = 3 ft = 0.914 m
final velocity, v = 0
a = g/2 = 4.9 m/s²
Use third equation of motion:
substitute the values to find the speed of train:
Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:
v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:
with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:
Furthermore, you use the relation between the friction force and the friction coefficient:
hence, the friction coefficient is 0.56