To solve for the magnitude of the force exerted on the swimmer by the water, we can use the following concepts from kinematics and Newton's second law of motion.

1. Kinematic Equation for Constant Acceleration:

First, we'll determine the acceleration. Since the swimmer is decelerating to a stop, the kinematic equation we can use is:

$v_f^2 = v_i^2 + 2a \cdot d$

where:

• $v_f = 0 \, \text{m/s}$ (final velocity, because the swimmer comes to rest),
• $v_i = 1.28 \, \text{m/s}$ (initial velocity),
• $a$ is the acceleration (which is negative because it's a deceleration),
• $d = 2.20 \, \text{m}$ (distance coasted).

We can solve for $a$:

$0 = (1.28)^2 + 2a \cdot 2.20$

$a = -\frac{(1.28)^2}{2 \cdot 2.20}$

2. Newton's Second Law:

Once we have the acceleration, we can use Newton’s second law to find the force. Newton’s second law is:

$F = m \cdot a$

where:

• $F$ is the force exerted by the water,
• $m = 48.5 \, \text{kg}$ (mass of the swimmer),
• $a$ is the acceleration (magnitude).

Let me calculate the values.The magnitude of the force exerted on the swimmer by the water is approximately 18.06 N. The negative sign indicates that the force is in the direction opposite to the swimmer's motion, but the magnitude is $18.06 \, \text{N}$.

The swimmer's acceleration is $-0.372 \, \text{m/s}^2$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the force change if the swimmer's mass were doubled?
2. What is the importance of constant acceleration in this problem?
3. How would the distance coasted affect the force exerted on the swimmer?
4. What would happen if the swimmer slowed down over a longer distance?
5. Can you apply this method to other cases of deceleration in fluids?

Tip: Always remember to keep track of the direction of forces and accelerations, especially in deceleration problems.