Question

Police investigators, examining the scene of an accident involving two cars, measure the 72m long skid marks of one of the cars with a mass of 2500kg. The coefficient of kinetic friction between rubber and pavement is about 0.80. Estimate the initial speed in mph of that car assuming a level road.

Answer 1

Answer on Question #42186 – Physics – Mechanics | Kinematics | Dynamics

Police investigators, examining the scene of an accident involving two cars, measure the 72m long skid marks of one of the cars with a mass of 2500kg. The coefficient of kinetic friction between rubber and pavement is about 0.80. Estimate the initial speed in mph of that car assuming a level road.

Solution:

$S = 72\mathrm{m}$ – traveled distance;

$m = 2500\mathrm{kg}$ – mass of the car;

$k = 0.80$ – coefficient of kinetic friction between rubber and pavement;

$V_{0}$ – initial speed of the car.

N – reaction force of the car;

Work-Energy theorem: work done by friction force if equal to change in kinetic energy of the car:

$$W _ {\text {f r i c t i o n}} = K E _ {\text {f i n a l}} - K E _ {\text {i n i t i a l}}$$ $$\mathrm {K E} _ {\text {f i n a l}} = 0 \text { (because final speed of the car is zero)}$$ $$\mathrm {W} _ {\text {f r i c t i o n}} = - \mathrm {K E} _ {\text {i n i t i a l}}$$ $$\mathrm {K E} _ {\text {i n i t i a l}} = \frac {\mathrm {m V} _ {0} ^ {2}}{2}$$ $$\mathrm {W} _ {\text {f r i c t i o n}} = - \mathrm {F} _ {\text {f r i c t i o n}} \cdot \mathrm {S} = - \mathrm {N k} \cdot \mathrm {S}$$

Work done by friction force has a minus sign, it shows that direction of the force is opposite to direction of motion (cars slows).

Second Newton's law along the Y-axis:

$$\mathrm {N} - \mathrm {m g} = 0$$ $$\mathrm {N} = \mathrm {m g} \Rightarrow$$ $$\mathrm {W} _ {\text {f r i c t i o n}} = - \mathrm {m g k} \cdot \mathrm {S}$$ $$(3) \text {a n d} (2) \text {i n} (1):$$ $$- \mathrm {m g k} \cdot \mathrm {S} = - \frac {\mathrm {m V} _ {0} ^ {2}}{2}$$ $$2 \mathrm {g k S} = \mathrm {V} _ {0} ^ {2}$$ $$V _ {0} = \sqrt {2 g k S} = \sqrt {2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 0 . 8 0 \cdot 7 2 m} = 3 3. 6 \frac {m}{s} = 7 5. 1 6 \mathrm {m p h}$$

Answer: initial speed is 75.16 mph.

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