Asked by Jerry
A bus moves from rest with a uniform acceleration of 2m/s² for the first 10s, it then accelerate at a uniform rate of 15m/s² for another 15s. it countinues at the same speed for 70s and finally comes to rest in 20s by uniform deceleration. find the total distance
Answers
Answered by
oobleck
accelerating: s = 1/2 at^2 = 100m
v = at = 20 m/s
s = 20t + 1/2 at^2 = 20*15 + 15/2 * 15^2 = 1987.5 m
then v = 20 + at = 245 m/s
constant: s = vt = 245*70 = 17150m
decelerating: a = -245/20 = -12.25 m/s^2
s = 245t + 1/2 at^2 = 245*20 - 6.125 * 20^2 = 2450m
total distance = 100+1987.5+17150+2450 = ______m
better double-check my arithmetic
v = at = 20 m/s
s = 20t + 1/2 at^2 = 20*15 + 15/2 * 15^2 = 1987.5 m
then v = 20 + at = 245 m/s
constant: s = vt = 245*70 = 17150m
decelerating: a = -245/20 = -12.25 m/s^2
s = 245t + 1/2 at^2 = 245*20 - 6.125 * 20^2 = 2450m
total distance = 100+1987.5+17150+2450 = ______m
better double-check my arithmetic
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