A driver of a car applied the brakes and barely avoided hitting an obstacle on the roadway. The driver left a skid mark of 28 m. Assuming that the coefficient of longitudinal deceleration is 0.41, determine whether the driver was in violation of the 70 km/h speed limit at that location if the driver was travelling
a. Uphill on a 3% incline
b. Downhill on a 3.3% incline
c. On a level roadway.
a. Uphill on a 3% incline
b. Downhill on a 3.3% incline
c. On a level roadway.
-
I'll answer c first:
coefficient = co
c) F = m.a
m.g.co = m.a
g. co = a
10. 0.41 = a
a = 4.1 m/sec^2
v = a.t
S = 1/2 a t^2
t = sqrt(2 S/a)
v = a.t = a sqrt(2 S/a) = sqrt(2Sa) = sqrt (2x28x4.1) = 15.15 m/sec^2
in km/h = 15.15 x 3600 / 1000 = 54.55 km/h
b) F = m.a
m.g.co. cos 3 - mg sin 3 = m.a
g.co. cos 3 - g sin 3 = a
a = 10 x 0.41 x 0.999 - 10 x 0.052 = 4.0959 - 0.52
a = 3.576 m/sec^2
v = sqrt(2Sa) = sqrt (2x28x3.576) = 14.15 m/sec^2
in km/h = 14.15 x 3600 / 1000 = 50.94 km/h
a) F = m.a
m.g.co. cos 3 + mg sin 3 = m.a
g.co. cos 3 + g sin 3 = a
a = 10 x 0.41 x 0.999 + 10 x 0.052 = 4.0959 + 0.52
a = 4.616 m/sec^2
v = sqrt(2Sa) = sqrt (2x28x4.616) = 16.078 m/sec^2
in km/h = 16.078 x 3600 / 1000 = 57.88 km/h
He didn't violate at all
coefficient = co
c) F = m.a
m.g.co = m.a
g. co = a
10. 0.41 = a
a = 4.1 m/sec^2
v = a.t
S = 1/2 a t^2
t = sqrt(2 S/a)
v = a.t = a sqrt(2 S/a) = sqrt(2Sa) = sqrt (2x28x4.1) = 15.15 m/sec^2
in km/h = 15.15 x 3600 / 1000 = 54.55 km/h
b) F = m.a
m.g.co. cos 3 - mg sin 3 = m.a
g.co. cos 3 - g sin 3 = a
a = 10 x 0.41 x 0.999 - 10 x 0.052 = 4.0959 - 0.52
a = 3.576 m/sec^2
v = sqrt(2Sa) = sqrt (2x28x3.576) = 14.15 m/sec^2
in km/h = 14.15 x 3600 / 1000 = 50.94 km/h
a) F = m.a
m.g.co. cos 3 + mg sin 3 = m.a
g.co. cos 3 + g sin 3 = a
a = 10 x 0.41 x 0.999 + 10 x 0.052 = 4.0959 + 0.52
a = 4.616 m/sec^2
v = sqrt(2Sa) = sqrt (2x28x4.616) = 16.078 m/sec^2
in km/h = 16.078 x 3600 / 1000 = 57.88 km/h
He didn't violate at all