Asked by Madison
A bicycle is moving at a constant speed of 12m/s as it covers a distance of 200m. it then applies its brakes and stops over the next 350m. how much time doe sthis total motion take?
Answers
Answered by
Steve
v = 12-at
The time necessary to come to a stop is when v=0, so a = 12/t
During that time t, the distance
s = 12t - 1/2 a t^2 = 350
1/2(12/t)t^2 - 12t + 350 = 0
6t-12t+350=0
6t=350
t = 58.33333 sec
a = 12/58.33333 = .2057m/s^2
Check the distance traveled:
s = 12(58.3333) - 1/2(.2057)*58.3333^2 = 350
So, the first 200m takes 200/12 = 16.6666s
the total time is 16.667 + 58.333 = 75 seconds
The time necessary to come to a stop is when v=0, so a = 12/t
During that time t, the distance
s = 12t - 1/2 a t^2 = 350
1/2(12/t)t^2 - 12t + 350 = 0
6t-12t+350=0
6t=350
t = 58.33333 sec
a = 12/58.33333 = .2057m/s^2
Check the distance traveled:
s = 12(58.3333) - 1/2(.2057)*58.3333^2 = 350
So, the first 200m takes 200/12 = 16.6666s
the total time is 16.667 + 58.333 = 75 seconds
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