A pizza boy on his way to delivering pizza gets into an accident. As he came aro

Question

A pizza boy on his way to delivering pizza gets into an accident. As he came around curve, he saw a 1.25 m^3 construction pallet sitting in the road. unable to stop in time, he hit struck the pallet, launching his magnetic “Pizza” roof sign in the air. The pizza boy claimed he was obeying the speed limit and wasn’t able to stop in time and now wants to sue. Public safety claim that the pizza delivery vehicle was traveling significantly faster than the curve.

Determine:

A. if the pizza boy would have had ample distance to bring his vehicle safely to a stop if he had obeyed the speed limit and, if possible.

B. the speed of the pizza delivery vehicle at the instant the delivery boy saw the pallet.

C. As a final step, please calculate the distance required to skid to a stop, when traveling at the speed limit, for a wet road. (Coefficient of friction for we surface = 0.4)

Data:

• The distance dtotal from the blind curve to the collision point is 28 m.

• The posted speed limit is 15 mph.

• Reaction time is 0.1438 seconds.

• The road is level with coefficient of friction of 0.863.

• The distance the car traveled, once the trailer was noted by the driver, includes both the distance traveled while the pizza boy was reacting to the situation and the distance traveled while skidding.

(dtotal = dreact + dskid)

• The car is 1.8 m high at the point where the magnetic pizza sign was attached.

• The magnetic pizza sign, which had been attached near the front of the car roof, landed 3.8 m from the collision point (assume kinetic friction and air resistance forces are negligible).

Explanation / Answer

speed limit = vl = 15 mph = 6.7056 m/s

distance treravelled in reaction time = dreact = 6.7056*0.1438 = 0.9643 m

acceleration while skidding = a = u*g = 0.863*9.81 = 8.46603 m/s^2

distance travelled while skidding = dskid = vl^2/2a = (6.7056^2)/(2*8.46603) = 2.6556 m

dtotal = dreact+ dskid = 3.6199 m

d total < 28

the boy had ample distance to bring his vehicle to a stop

+++++++++++++

(B)

let v be the speed of the vehicle

dreact = v*0.1438

dskid = v^2/(2*8.46603)

d total = dreact + dskid

28 = v*0.1438 + v^2/(2*8.46603)

v = 20.6 m/s    <------answer

++++++++++++

(C)

dreact = v*treact = 20.6*0.1438 = 2.96228 m

the final velocity = 0

acceleration = 0.4*9.81 = 3.924 m/s^2

dskid = v^2/(2*a) = 20.6^2/(2*3.924) = 54.0723 m

dtotal = 2.96228+54.0723 = 57.03458 m <------answer

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