Michael took the return trip at a velocity 33.75 miles per hour.
<h3>How fast did Michael drive in his return trip?</h3>
Let suppose that Michael drove in <em>straight line</em> road and at <em>constant</em> velocity. Therefore, the speed of the vehicle (v), in miles per hour, can be defined as distance traveled by the vehicle (d), in miles, divided by travel time (t), in hours.
First trip
45 = s / 3 (1)
Second trip
v = s / 4 (2)
By (1) and (2):
45 · 3 = 4 · v
v = 33.75 mi / h
Michael took the return trip at a velocity 33.75 miles per hour.
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Answer:
Third
Step-by-step explanation:
Acceleration is the rate of change of the rate of change of position
In the third, the rate of change is negative, and the rate of change of that is decreasing as time passes.
Step-by-step explanation:
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