A car travels along a straight line at a constant speed of 41.5 mi/h for a dista

Question

A car travels along a straight line at a constant speed of 41.5 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 35.0 mi/h.
(a) What is the constant speed with which the car moved during the second distance d?
(b) Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip?
(c) What is the average speed for this new trip?

Explanation / Answer

a)
E1: v1=43.0 mi/h
E2: d = v1*t1 [1st distance, constant speed v1]
E3: d = v2*t2 [2nd distance, another constant speed v2]
E4: v3=33.5 mi/h
E5: 2d = v3*(t1+t2) [double distance, average velocity v3]

Combine E2 and E3 and calculate t2
v2*t2 = v1*t1
E6: t2 = v1*t1/v2

In E5, substitute t2 with t2 from E6 and d with d from E2
2*v1*t1 = v3*(t1 + v1*t1/v2)
2*v1*t1 = v3*t1 + v3*v1*t1/v2
2*v1*t1 - v3*t1 = v3*v1*t1/v2
(2*v1-v3)*t1 = v3*v1*t1/v2
2*v1-v3 = v3*v1/v2
or
E7: v2 = v3*v1/(2*v1-v3)

In E7 substitute v1 with v1 from E1 and v3 with v3 from E4
v2 = 33.5*43 / (2*43 - 33.5)
v2 = 27.438095 mi/h
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b)
There are 2 options:
b1) if you take distance d as a "scalar" then you traveled d+d = 2d and the velocity is the same as in a)
b2) if you take distance d as a "vector" then you traveled d-d = 0 and the velocity is 0
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c)
English is my second language, but what is the difference between "speed" and "velocity"?
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