# An airplane moving with a velocity equal to 300 mph when the velocity was increased with uniform acceleration to 600 mph in 5 minutes.?

Update:

Please calculate the number of miles the airplane traveled during that interval of time. Please show work.

Relevance

300 mph = 300/60 miles per minute = 5 miles per minute

600 mph = 600/60 miles per minute = 10 miles per minute

v(5) = 10

v(0) = 5

(10 - 5) / (5 - 0) = 5/5 = 1

v(t) = t + 5

Integrate

s(t) = (1/2) * t^2 + 5t + C

s(5) - s(0) =>

(1/2) * 5^2 + 5 * 5 - (1/2) * 0^2 - 5 * 0 =>

(1/2) * 25 + 25 =>

37.5

37.5 miles

Or we could avoid changing the mph to miles per minute by converting 5 minutes to 1/12 hours

v(1/12) = 600

v(0) = 300

(600 - 300) / (1/12 - 0) =>

300 / (1/12) =>

3600

v(t) = 3600 * t + 300

s(t) = 1800 * t^2 + 300 * t + C

s(1/12) - s(0) =>

1800 * (1/12)^2 + 300 * (1/12) - 1800 * 0^2 - 300 * 0 =>

1800 * (1/144) + 25 =>

18 * 100 / (36 * 4) + 25 =>

100 / (2 * 4) + 25 =>

50 / 4 + 25 =>

12.5 + 25 =>

37.5

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• v = at + v0 [v:mph, t:h, a:m/h^2]

600 = a * (5/60) + 300

a = 3600 mile/h^2

x = 1/2 * at^2 + v0 * t + x0

x = 1/2 * 3600 * (5/60)^2 + 300 * (5/60) + 0

x = 37.5 miles

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• In uniform acceleration, we can find the average velocity by taking the average of the initial and final velocity.

Assuming it is a straight line motion, multiply that by the time (1/12 hr) to find distance.

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