Question #47440

Sally travels by car from one city to another. She drives for 28.0 min at 81.0 km/h, 33.0 min at 36.0 km/h, and 32.0 min at 69.0 km/h, and she spends 12.0 min eating lunch and buying gas.
(a) Determine the average speed for the trip.

(b) Determine the total distance traveled.
1

Expert's answer

2014-10-02T09:50:20-0400

Answer on Question #47440, Physics, Mechanics - Kinematics - Dynamics

Sally travels by car from one city to another. She drives for 28.0 min at 81.0 km/h, 33.0 min at 36.0 km/h, and 32.0 min at 69.0 km/h, and she spends 12.0 min eating lunch and buying gas.

(a) Determine the average speed for the trip.

(b) Determine the total distance traveled.

Solution:

By the definition, average speed is:


vav=Stotalttotalv_{av} = \frac{S_{total}}{t_{total}}


To find time in hours:


th=tmin60t_{h} = \frac{t_{min}}{60}


Traveled distance:


S=vtS = vt


So, average speed is:


vav=v1t160+v2t260+v2t260t160+t260+t360+t460vav=81.0kmh28.060h+36.0kmh33.060h+69.0kmh32.060h28.060h+33.060h+36.060h+12.060h52.0kmh\begin{aligned} v_{av} &= \frac{v_{1} \frac{t_{1}}{60} + v_{2} \frac{t_{2}}{60} + v_{2} \frac{t_{2}}{60}}{\frac{t_{1}}{60} + \frac{t_{2}}{60} + \frac{t_{3}}{60} + \frac{t_{4}}{60}} \\ v_{av} &= \frac{81.0 \frac{km}{h} \cdot \frac{28.0}{60} h + 36.0 \frac{km}{h} \cdot \frac{33.0}{60} h + 69.0 \frac{km}{h} \cdot \frac{32.0}{60} h}{\frac{28.0}{60} h + \frac{33.0}{60} h + \frac{36.0}{60} h + \frac{12.0}{60} h} \approx 52.0 \frac{km}{h} \end{aligned}


And total traveled distance


s=v1t160+v2t260+v2t260=81.0kmh28.060h+36.0kmh33.060h+69.0kmh32.060h=94.4kms = v_{1} \frac{t_{1}}{60} + v_{2} \frac{t_{2}}{60} + v_{2} \frac{t_{2}}{60} = 81.0 \frac{km}{h} \cdot \frac{28.0}{60} h + 36.0 \frac{km}{h} \cdot \frac{33.0}{60} h + 69.0 \frac{km}{h} \cdot \frac{32.0}{60} h = 94.4 \, km


**Answer:** Average speed: vav52.0kmhv_{av} \approx 52.0 \frac{km}{h}

Traveled distance is s=94.4kms = 94.4 \, km

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