A power boat of mass 465 kg is cruising at a constant speed of 8.9 m/s. The propeller provides a drive force of 765 N. The driver of the boat shuts off the engine, and the boat coasts to a halt. Assume-contrary to fact-that the resistive force due to the water is constant, independent of the boat's speed.
(a) How far does the boat coast?
(b) How much time does it take for the boat to come to rest after the engine is turned off?
friction is 765N.
acceleration during slowing is F/mass=765/m
Vf^2=Vi^2+2ad where a= -765/m solve for d.
average velocity= 8.9m/s
time= distance/8.9m/s
To find the distance the boat coasts, we can use the equation:
\(F = ma\)
where F is the net force, m is the mass, and a is the acceleration.
(a) In this case, the net force is equal to the resistive force due to the water. The propeller force is no longer present since the engine is turned off. Given that the resistive force is constant, we have:
\(F = F_{\text{resistive}} = -F_{\text{propeller}}\)
Substituting the values, we have:
\(F_{\text{resistive}} = -765 \, \text{N}\)
Since we also have \(F = ma\), we can rearrange the equation to solve for acceleration:
\(a = \frac{F}{m}\)
Substituting the values, we get:
\(a = \frac{-765 \, \text{N}}{465 \, \text{kg}}\)
Simplifying, we find:
\(a = -1.645 \, \text{m/s}^2\)
The acceleration is negative because it is in the opposite direction of motion.
To find the distance, we can use the equation:
\(d = v_0t + \frac{1}{2}at^2\)
Where:
\(d\) is the distance
\(v_0\) is the initial velocity (8.9 m/s)
\(t\) is the time
\(a\) is acceleration
Since the boat coasts to a halt, the final velocity \(v\) will be 0. We can set \(v = 0\) and solve for \(t\).
We have \(v = v_0 + at\), which becomes:
\(0 = 8.9 \, \text{m/s} -1.645 \, \text{m/s}^2 \cdot t\)
Solving for \(t\), we find:
\(t = \frac{8.9 \, \text{m/s}}{1.645 \, \text{m/s}^2}\)
Calculating, we find:
\(t \approx 5.407 \, \text{s}\)
Now we can substitute the values back into the equation for distance:
\(d = (8.9 \, \text{m/s}) \cdot (5.407 \, \text{s}) + \frac{1}{2} (-1.645 \, \text{m/s}^2)(5.407 \, \text{s})^2\)
Calculating, we find:
\(d \approx 24.28 \, \text{m}\)
Therefore, the boat coasts approximately 24.28 meters.
(b) The time it takes for the boat to come to rest after the engine is turned off is approximately 5.407 seconds.
To answer these questions, we need to use Newton's second law of motion and the equation for calculating distance and time.
First, let's calculate the acceleration of the boat when the engine is turned off. We can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given that the propeller provides a drive force of 765 N and the boat's mass is 465 kg, we can calculate the acceleration:
Net force = mass × acceleration
765 N = 465 kg × acceleration
Now, solve for acceleration:
acceleration = 765 N / 465 kg
acceleration ≈ 1.65 m/s²
(a) To find the distance the boat coasts, we can use the equation for calculating distance:
distance = initial velocity × time + 0.5 × acceleration × time²
Since the boat is initially traveling at a constant speed of 8.9 m/s and it comes to a stop, the final velocity is 0. Therefore, we can use the equation:
0 = 8.9 m/s + 0.5 × 1.65 m/s² × time²
Solve for time:
4.45 m/s = 0.5 × 1.65 m/s² × time²
time² = (4.45 m/s) / (0.5 × 1.65 m/s²)
time² ≈ 1.35 s
Now, we can use the value of time to calculate the distance:
distance = 8.9 m/s × 1.35 s + 0.5 × 1.65 m/s² × (1.35 s)²
distance ≈ 6.0 m
Therefore, the boat coasts approximately 6.0 meters.
(b) To find the time it takes for the boat to come to rest, we can simply use the value of time we calculated above:
time ≈ 1.35 s
Therefore, it takes approximately 1.35 seconds for the boat to come to rest after the engine is turned off.