Squash ball acceleration

Here are the solutions to the numerical questions:

1. Squash ball acceleration

• Initial velocity ($v_i$): $15\, \text{m/s}$ west (let's take west as negative, so $v_i = -15\, \text{m/s}$)
• Final velocity ($v_f$): $25\, \text{m/s}$ east (let's take east as positive, so $v_f = +25\, \text{m/s}$)
• Time ($t$): $0.10\, \text{s}$

Step 1: Use the formula for acceleration.

$$a = \frac{v_f - v_i}{t}$$

Step 2: Substitute the given values.

$$a = \frac{25\, \text{m/s} - (-15\, \text{m/s})}{0.10\, \text{s}}$$

Step 3: Calculate the acceleration.

$$a = \frac{40\, \text{m/s}}{0.10\, \text{s}}$$ $$a = 400\, \text{m/s}^2$$

The positive sign indicates the acceleration is in the eastward direction.
The vector acceleration is $\boxed{\text{400 m/s}^2 \text{ east}}$.

2. Golf ball displacement

• Initial velocity ($v_i$): $25\, \text{m/s}$
• Acceleration ($a$): $-20\, \text{m/s}^2$ (deceleration)
• Time ($t$): $2.0\, \text{s}$

Step 1: Use the kinematic equation for displacement.

$$s = v_i t + \frac{1}{2} a t^2$$

Step 2: Substitute the given values.

$$s = (25\, \text{m/s})(2.0\, \text{s}) + \frac{1}{2} (-20\, \text{m/s}^2)(2.0\, \text{s})^2$$

Step 3: Calculate the displacement.

$$s = 50\, \text{m} + \frac{1}{2} (-20\, \text{m/s}^2)(4.0\, \text{s}^2)$$ $$s = 50\, \text{m} - 40\, \text{m}$$ $$s = 10\, \text{m}$$

The displacement is $\boxed{\text{10 m}}$.

3. Bullet exit speed

• Displacement ($s$): $0.750\, \text{m}$
• Acceleration ($a$): $5.35 \times 10^5\, \text{m/s}^2$
• Initial speed ($v_i$): $0\, \text{m/s}$ (starts from rest)

Step 1: Use the kinematic equation relating final velocity, initial velocity, acceleration, and displacement.

$$v_f^2 = v_i^2 + 2as$$

Step 2: Substitute the given values.

$$v_f^2 = (0\, \text{m/s})^2 + 2(5.35 \times 10^5\, \text{m/s}^2)(0.750\, \text{m})$$

Step 3: Calculate $v_f^2$.

$$v_f^2 = 8.025 \times 10^5\, \text{m}^2/\text{s}^2$$

Step 4: Calculate $v_f$.

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