Squash ball acceleration

Here are the solutions to the numerical questions: 1. Squash ball acceleration Initial velocity (v_i): 15\, m/s west (let's take west as negative, so v_i = -15\, m/s) Final velocity (v_f): 25\, m/s east (let's take east as positive, so v_f = +25\, m/s) Time (t): 0.10\, s Step 1: Use the formula for acceleration. a = (v_f - v_i)/(t) Step 2: Substitute the given values. a = 25\, m/s - (-15\, m/s)0.10\, s Step 3: Calculate the acceleration. a = 40\, m/s0.10\, s a = 400\, m/s^2 The positive sign indicates the acceleration is in the eastward direction. The vector acceleration is 400 m/s^2 east. 2. Golf ball displacement Initial velocity (v_i): 25\, m/s Acceleration (a): -20\, m/s^2 (deceleration) Time (t): 2.0\, s Step 1: Use the kinematic equation for displacement. s = v_i t + (1)/(2) a t^2 Step 2: Substitute the given values. s = (25\, m/s)(2.0\, s) + (1)/(2) (-20\, m/s^2)(2.0\, s)^2 Step 3: Calculate the displacement. s = 50\, m + (1)/(2) (-20\, m/s^2)(4.0\, s^2) s = 50\, m - 40\, m s = 10\, m The displacement is 10 m. 3. Bullet exit speed Displacement (s): 0.750\, m Acceleration (a): 5.35 × 10^5\, m/s^2 Initial speed (v_i): 0\, 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\, m/s)^2 + 2(5.35 × 10^5\, m/s^2)(0.750\, m) Step 3: Calculate v_f^2. v_f^2 = 8.025 × 10^5\, m^2/s^2 Step 4: Calculate v_f. v_f = sqrt(8.025 × 10^5\, m)^2/