Asked by Evan
A motorcycle racer falls off their motorbike while traveling at 102 km/hr and skids in a straight line down the track. If the friction with the track decelerates the rider at a uniform rate of 4.6 m/s/s, how far (what displacement) would the rider travel before coming to a stop?
Answers
Answered by
oobleck
102 km/hr = 28.33 m/s
Now, as usual,
v = 28.33 - 4.6t
find t when v=0, and then
s = 28.33t - 2.3t^2
Now, as usual,
v = 28.33 - 4.6t
find t when v=0, and then
s = 28.33t - 2.3t^2
Answered by
Evan
I'm still lost on what the answer is?
how did you get 28.33 m/s?
how did you get 28.33 m/s?
Answered by
bobpursley
102 km/hr = 28.33 m/s from
102km/hr*1000m/km*1hr/3600sec= 28.33 m/s
If you are serious with this question, you need some tutoring on unit conversion very quickly.
102km/hr*1000m/km*1hr/3600sec= 28.33 m/s
If you are serious with this question, you need some tutoring on unit conversion very quickly.
Answered by
Yashvi
Hi, im just curious and would appreciate it if someone replied. Would the answer be 87.24 m?
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