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
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.)
7 answers
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
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
ignore the very last "the acceleration is 9.33"
pressed "post answer" too soon
so the acceleration is a = 13.59 ft/s^2
and
v = 13.59t + 32.2667
a = 13.59 ft/s^2
pressed "post answer" too soon
so the acceleration is a = 13.59 ft/s^2
and
v = 13.59t + 32.2667
a = 13.59 ft/s^2
4.14 m/s^2 = 13.6 ft/s^2 :)
what's wrong with 28/3 mi/hr/s ? :-)
Nothing. I use SCI unless ordered otherwise :)
I've tried your answers to this question but they were all marked wrong.
Thanks anyway!
Thanks anyway!