Asked by Dian
A jogger is moving at 5 m/s as she approaches a busy street. She needs to stop in 2 seconds in order to stay safe. What average deceleration must she have in order to stop in time?
Answers
Answered by
Erica
using the kinematic equation: Vf = Vo + at
Vf = final velocity = 0m/s because she comes to a stop
Vo = initial velocity = 5m/s as she is jogging
a = acceleration (or in this case deceleration)
t = time elapsed = 2 s
so...
Vf = Vo + at
0m/s = 5m/s + a * 2s
-5m/s = a * 2s
(-5m/s)/2s = a
-2.5 m/s^2 = a
because the acceleration ends up being negative it is understood to be deceleration.
Vf = final velocity = 0m/s because she comes to a stop
Vo = initial velocity = 5m/s as she is jogging
a = acceleration (or in this case deceleration)
t = time elapsed = 2 s
so...
Vf = Vo + at
0m/s = 5m/s + a * 2s
-5m/s = a * 2s
(-5m/s)/2s = a
-2.5 m/s^2 = a
because the acceleration ends up being negative it is understood to be deceleration.
Answer
help me now
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.