You are driving at a constant speed V 0 on a straight highway that passes under

Question

You are driving at a constant speed V0 on a straight highway that passes under an overpass. At time t = 0, when your car is a distance d from the overpass, a prankster on the overpass releases a water balloon from rest from a height h above the highway below.

Answer the following questions. You should treat the car as a point particle, ignoring its height, length, and width.

(a) Suppose that you don't see the water balloon until it explodes on your windshield. At what speed V0 were you originally traveling, given that the water balloon strikes your car? Express your answer in terms of d, h, and any relevant constants. Simplify your answer as much as possible.

(b) Now suppose that you see the prankster release the water balloon at t = 0, and you slam on your brakes at time tb in an attempt to avoid being hit. (Your reaction time is tb > 0.) Your car slows down with a constant acceleration of magnitude as. The balloon hits the highway right in front of your car at the instant your car comes to rest. At what speed V0 were you originally traveling, given that the water balloon just barely misses your car? Express your answer in terms of d, h, as, tb, and any relevant constants. Simplify your answer as much as possible.

(c) Let Va be the speed you found in part (a), and let Vb be the speed you found in part (b). Which of these two speeds is larger? Justify your answer mathematically, with comments as necessary.

Explanation / Answer

a)d=Vo*t (1)

h=0.5*g*t^2. (2)

eliminating t from both equation Vo=d*(g/2h)^0.5.

b)time taken by ballon to reach the ground(t)=(2h/g)^0.5. (from (2))

also, d=Vo*tb+ Vo*(t-tb) - 0.5*as*(t-tb)^2.

So Vo= (d + 0.5as*(t-tb)^2)/t

=(d + 0.5as*((2h/g)^0.5-tb)^2)/(2h/g)^0.5.

c) Vb is larger

Vb= d*(g/2h)^0.5 + 0.5as*((2h/g)^0.5-tb)^2)/(2h/g)^0.5.

Vb=Va + C (where C is a positive constant).

Hence Vb is larger

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