What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

Three cars are driving at 27.0m/s along the road shown in the figure (Figure 1) . Car B is at the bottom of the hill and car C is at the top. Suppose each car suddenly brakes hard and starts to skid. Assume μk=1.00.

Answer 1

This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

a) In this one we don’t need to worry about the normal acceleration since there is no curve.

at = uk*Fn

Fn = mg = 9.81m

at = 1*9.81m = -9.81*m

this is considered negative since the forward motion of the car is in the positive direction.

b) Now we will find the normal acceleration caused by the curve.

an = v^2/r = 27^2/200 = 0.135 m/s^2

Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

at = uk*Fn = 1*9.945m = -9.945m

c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

an = 0.135

Fn = (9.81 – 0.135)m = 9.675m

at = uk*Fn = -9.675m

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