A power boat of mass 495 kg is cruising at a constant speed of 8.9 m/s. The propeller provides a drive force of 780 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?
m
(b) How much time does it take for the boat to come to rest after the engine is turned off?

a) Set F*X equal to the initial kinetic energy and solve for X. X is the distance and f is the friction force, which is here assumed to equal the propeller force while running at full speed, 780 N

b) Divide X by the average speed, V/2.

To find the distance the boat coasts before coming to a halt, we can use the equation of motion:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s, as the boat comes to a halt)
u = initial velocity (8.9 m/s)
a = acceleration (which is the net force divided by the mass)
s = distance

(a) Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Since the boat is decelerating, the net force will be the resistive force due to the water, which is equal in magnitude to the propeller force (780 N).

Therefore, the acceleration can be found using Newton's second law:

F = ma
780 N = (495 kg) * a

Solving for a, we find:

a = 780 N / 495 kg
a ≈ 1.576 m/s^2

Now we can substitute the values into the distance equation:

s = (0^2 - (8.9 m/s)^2) / (2 * 1.576 m/s^2)
s = (-79.21 m^2/s^2) / 3.152 m/s^2
s ≈ -25.11 m^2/s^2 / m/s^2
s ≈ -25.11 m

However, distance cannot be negative, so we take the magnitude:

s ≈ 25.11 m

Therefore, the boat coasts for approximately 25.11 meters before coming to a halt.

(b) To find the time it takes for the boat to come to rest, we can use the equation:

v = u + at

where:
v = final velocity (0 m/s)
u = initial velocity (8.9 m/s)
a = acceleration (-1.576 m/s^2, negative because the boat is decelerating)
t = time

Rearranging the equation, we get:

t = (v - u) / a
t = (0 m/s - 8.9 m/s) / (-1.576 m/s^2)
t = (-8.9 m/s) / (-1.576 m/s^2)
t ≈ 5.65 s

Therefore, it takes approximately 5.65 seconds for the boat to come to rest after the engine is turned off.

To solve this problem, we need to consider the forces acting on the boat and apply the laws of motion.

(a) How far does the boat coast?

The motion of the boat can be described by Newton's second law: F = ma, where F is the net force acting on the boat, m is the mass of the boat, and a is the acceleration.

Initially, when the engine is on, the propeller force (780 N) and the resistive force due to the water (let's call it R) act on the boat. So the net force is:

F_net = Propeller force - Resistive force = 780 N - R

Since the boat is cruising at a constant speed of 8.9 m/s, the net force is zero. Therefore:

0 = 780 N - R

R = 780 N

When the engine is turned off, the only force acting on the boat is the resistive force due to the water. According to the problem statement, this resistive force is constant.

Now, we apply Newton's second law to determine the acceleration of the boat after the engine is turned off. We assume that the opposing resistive force is equal to the force required to decelerate the boat to a stop.

F_net = ma

Resistive force (R) = ma

780 N = 495 kg * a

a = 780 N / 495 kg ≈ 1.576 m/s²

Since we know the initial speed (v₀ = 8.9 m/s) and the acceleration (a = -1.576 m/s²), we can use the kinematic equation to determine the distance traveled while decelerating to zero speed:

v² = v₀² + 2ad

0² = (8.9 m/s)² + 2(-1.576 m/s²) * d

0 = 79.21 m²/s² - 3.152 m/s² * d

3.152 m/s² * d = 79.21 m²/s²

d ≈ 25.12 m

Therefore, the boat coasts for approximately 25.12 meters.

(b) How much time does it take for the boat to come to rest after the engine is turned off?

To find the time, we can use the equation:

v = v₀ + at

where v is the final velocity (0 m/s), v₀ is the initial velocity (8.9 m/s), a is the acceleration (-1.576 m/s²), and t is the time.

0 = 8.9 m/s + (-1.576 m/s²) * t

-8.9 m/s = -1.576 m/s² * t

t ≈ 5.65 s

Therefore, it takes approximately 5.65 seconds for the boat to come to rest after the engine is turned off.