Asked by mssailormouth
A car accelerates from 0.0 to 33 m/s in 6.0 seconds. What is the magnitude of the acceleration?
I got 198 but the answer is 5.5?? Not understanding this.
I got 198 but the answer is 5.5?? Not understanding this.
Answers
Answered by
oobleck
acceleration is the rate of change of velocity
a = ∆v/∆t = (33-0)m/s / (6-0)s = 5.5 m/s^2
how did you get 198?
33*6 = 198, but that is m/s * s = meters
That is the distance covered in 6 seconds at a constant speed of 33 m/s
A good way to avoid that kind of mistake is to keep track of the units. If you don't wind up with m/s^2 you have done something wrong.
a = ∆v/∆t = (33-0)m/s / (6-0)s = 5.5 m/s^2
how did you get 198?
33*6 = 198, but that is m/s * s = meters
That is the distance covered in 6 seconds at a constant speed of 33 m/s
A good way to avoid that kind of mistake is to keep track of the units. If you don't wind up with m/s^2 you have done something wrong.
Answered by
Reiny
v = at + c
when t = 0, v = 0 , thus c = 0
so we have <b>v = at</b>
when t = 0 , v = 0
when t = 6, v = 33
33 = 6a
a = 33/6 = 5.5
when t = 0, v = 0 , thus c = 0
so we have <b>v = at</b>
when t = 0 , v = 0
when t = 6, v = 33
33 = 6a
a = 33/6 = 5.5
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