There\'s a stretch of road that uses cameras to compute the average velocity of

Question

There's a stretch of road that uses cameras to compute the average velocity of cars. The cameras are 4 miles apart and the speed limit is 50mph. You are traveling at 50mph as you pass one camera and 4 minutes later you pass the second camera at 50mph. You are issued a speeding ticket.

a) What was your average velocity during the time you were between the 2 cameras?

b) Were you speeding at some time that you were between the 2 cameras? State the name of any theorem from Calculus you used to come to this conclusion.

Explanation / Answer

a)

displacement = 4 miles

time = 4 min = 4/60 hr =1/15 hr

average velocity = displacement/time

= 4 miles / (1/15 hr)

= 60 mph

Answer: 60 mph

b)

we can use mean value theorem which states that

f’(c) = (f(b) - f(a)) / (b-a)

b-a is time difference

f(b)-f(a) is difference in distance

f’(c) is speed at some point between a and b

f’(c) = (4 miles)/ (1/15 hr)

f’(c) = 60 mph

So, at some point speed was 60 mph

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