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?
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.
To find the correct answer, let's go through the steps again.
Step 1: Write down the known values:
Initial velocity (Vo) = 33.7 m/s
Acceleration (a) = -2.9 m/s^2 (since it's decelerating, the acceleration is negative)
Distance to the buoy (x) = 93.3 m
Step 2: Use the equation of motion:
x = Vot + (1/2)at^2
Plugging in the values we have:
93.3 = 33.7t + (1/2)(-2.9)t^2
Step 3: Rearrange the equation to solve for time:
This equation is a quadratic equation, which means it can have two solutions. Since we're interested in the positive time it takes to reach the buoy, we discard the negative solution.
To find the time, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, the equation is:
(1/2)(-2.9)t^2 + 33.7t - 93.3 = 0
Plug in the values of a, b, and c into the formula, and you should get two possible values for t. Choose the positive value.
t = (-33.7 ± √(33.7^2 - 4(1/2)(-2.9)(-93.3))) / (2(1/2)(-2.9))
Alternatively, you can simplify the equation before solving for t:
t^2 - 23.15t + 64.275 = 0
Solve this quadratic equation correctly and you should find the correct value for 't'.