A speedboat moving at 30 m/s approaches a no-wake buoy marker 100 m ahead. The pilot
slows the boat with a constant acceleration of 3.50 m/s2
by reducing the throttle.
v = Vi + a t
v = 30 - 3.5 t
when does it stop ?
t = 30/3.5 = 8.57 s
average speed during stop = 15 m/s
so
15 *8.57 = 129 meters, heavens to murgatroid, ran over the buoy
so how fast was he going when he hit the buoy>
d = 100 = 30 t - (1/2) 3.50 t^2
100 = 30 t - 1.75 t^2
t = 4.5 or 12.6 seconds
use 4.5 seconds, 12.6 is after you run over it and come back and hit it again in reverse
then v = 30 - 3.5*4.5 = 30 - 15.75= 14.25 m/s as he runs over the buoy
To find the time it takes for the speedboat to reach the no-wake buoy marker, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s at the marker)
u = initial velocity (30 m/s)
a = acceleration (-3.50 m/s^2, negative because the boat is slowing down)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values:
t = (0 - 30) / (-3.50)
t = 8.57 seconds
Therefore, it takes approximately 8.57 seconds for the speedboat to reach the no-wake buoy marker.
To determine the time it takes for the speedboat to reach the no-wake buoy marker, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
t = time
a = acceleration
In this case, the speedboat is moving at an initial velocity of 30 m/s and has a constant deceleration of -3.50 m/s² (since slowing down is considered to be a negative acceleration).
We need to determine the time it takes for the boat to reach the buoy, which is 100 m ahead. So, the distance (s) is 100 m.
Using the equation, we rearrange it to solve for time (t):
100 = 30t + (1/2)(-3.50)t^2
Next, we simplify the equation:
100 = 30t - 1.75t^2
Rearranging the equation to be in standard quadratic form:
1.75t^2 - 30t + 100 = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case:
a = 1.75
b = -30
c = 100
Let's calculate the values:
t = [ -(-30) ± √((-30)^2 - 4(1.75)(100)) ] / (2(1.75))
Simplifying further:
t = (30 ± √(900 - 700)) / (3.5)
t = (30 ± √200) / (3.5)
Now, we calculate the time using both the positive and negative square root values:
t1 = (30 + √200) / 3.5 ≈ 9.30 seconds
t2 = (30 - √200) / 3.5 ≈ -3.20 seconds
Since time cannot be negative in this context, we discard the negative value. Therefore, the time it takes for the speedboat to reach the no-wake buoy marker is approximately 9.30 seconds.