Asked by Michelle
A speed boat moving at 33.7 m/s approaches a
no-wake buoy marker 93.3 m ahead. The pilot
slows the boat with a constant deceleration of
2.9 m/s
2
by reducing the throttle.
How long does it take the boat to reach the
buoy?
current work done:
x = Vot + (1/2)at^2
93.3 = 0(t) + (1/2)(-2.9)t^2
t = 8.0215
however when I submit that its wrong.
What do I do?
no-wake buoy marker 93.3 m ahead. The pilot
slows the boat with a constant deceleration of
2.9 m/s
2
by reducing the throttle.
How long does it take the boat to reach the
buoy?
current work done:
x = Vot + (1/2)at^2
93.3 = 0(t) + (1/2)(-2.9)t^2
t = 8.0215
however when I submit that its wrong.
What do I do?
Answers
Answered by
Steve
I think you misunderstand Vo. If there were no acceleration, you'd have x=Vo*t, and it would not slow down at all.
The equation you have is right, but Vo = 33.7, its starting speed.
93.3 = 33.7t + 1/2 (-2.9)t^2
t = 3.2126 or 20.029
Why two answers? Which is right?
Think of the situation. After 3.2126 seconds the boat has reached the buoy.
But it has not stopped. A while later it stops and then comes back again (acceleration keeps reducing the velocity until it becomes negative). After 20.029 seconds, the boat is back at the buoy.
The equation you have is right, but Vo = 33.7, its starting speed.
93.3 = 33.7t + 1/2 (-2.9)t^2
t = 3.2126 or 20.029
Why two answers? Which is right?
Think of the situation. After 3.2126 seconds the boat has reached the buoy.
But it has not stopped. A while later it stops and then comes back again (acceleration keeps reducing the velocity until it becomes negative). After 20.029 seconds, the boat is back at the buoy.
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