# calculus help

What constant acceleration is required to increase the speed of a car from 22 mi/h to 50 mi/h in 3 s? (Round your answer to two decimal places.)

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1. 22 mph = 9.83 m/s
50 mph = 22.35 m/s

a = change in velocity/time = (22.35 -9.81)/3
= 4.18 m/s^2 or about half g

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2. If the acceleration is a
then
v = at + c , where c is a constant:

when t=0 , v = 22 mi/h = 32.2666.. ft/s
32.266.. = a(0) + c
c = 32.2666..

when t = 3s , v = 50 mi/h = 73.333.. ft/s
73.333.. = a(3) + 32.26666...
3a = 41.06666..
a = 13.58888..

the acceleration is 13.89 ft/s^2

the accelereation is 9.33

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3. ignore the very last "the acceleration is 9.33"

so the acceleration is a = 13.59 ft/s^2
and
v = 13.59t + 32.2667

a = 13.59 ft/s^2

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4. 4.14 m/s^2 = 13.6 ft/s^2 :)

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5. what's wrong with 28/3 mi/hr/s ? :-)

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6. Nothing. I use SCI unless ordered otherwise :)

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7. I've tried your answers to this question but they were all marked wrong.

Thanks anyway!

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