A car drives along a road on a stormy night. Darkness and heavy rain has reduced
ID: 1772639 • Letter: A
Question
A car drives along a road on a stormy night. Darkness and heavy rain has reduced the visibility to d=15 m. We will examine whether the driver can brake in time to avoid hitting a fallen tree which blocks the road. The kinetic coefficient of friction between the tires and the road surface is µk=0.4 when the brakes are locked. The driver’s reaction time is Dt=1 s and his reaction to seeing the fallen tree is to lock his brakes.
(a) Calculate the maximum safe speed that allows the driver to avoid colliding with the tree if the road is horizontal.
(b) Calculate the maximum safe speed if the car is going down hill on a road which is inclined by an angle or theta=15 degrees with respect to the horizontal direction.
Explanation / Answer
Take Vmax is maximum speed of the vehicle.
(A)
Write the expression for acceleration of the car.
a = -g*mue_k
= -9.8*0.4
= -3.92 m/s^2
Calculate the distance travelled in reaction time.
d1 = delta_t*Vmax
= 1*Vmax
Suppose d2 is the remainig distance.
Consider following equation of motion.
v^2 - u^2 = 2*a*s
0^2 - Vmax^2 = 2*a*d2
==> d2 = -Vmax^2/(2*a)
= -Vmax^2/(2*(-3.92))
= 0.12755*Vmax^2
d = d1 + d2
15 = Vmax + 0.12755*Vmax^2
Solve above equation for Vmax
Vmax = 7.6 m/s
(B)
Consider the following expression of acceleration of car on inclined road.
a = g*sin(15) -g*mue_k*cos(15)
= 9.8*sin(15) -9.8*0.4*cos(15)
= -1.25 m/s^2
Calculate the distance travelled in reaction time.
d1 = delta_t*Vmax
= 1*Vmax
Suppose d2 is the remainig distance.
Consider following equation of motion.
v^2 - u^2 = 2*a*s
0^2 - Vmax^2 = 2*a*d2
==> d2 = -Vmax^2/(2*a)
= -Vmax^2/(2*(-1.25))
= 0.4*Vmax^2
d = d1 + d2
15 = Vmax + 0.4*Vmax^2
Solve for Vmax
Vmax = 5 m/s
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