This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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b) i) Determine its velocity after 1.5 seconds. Step 1: Identify the given values and the formula. Initial velocity u = 10 \, m/s. Acceleration a = -2.5 \, m/s^2 (deceleration). Time t = 1.5 \, s. We use the kinematic equation: v = u + at. Step 2: Substitute the values into the formula and calculate the final velocity. v = 10 \, m/s + (-2.5 \, m/s^2)(1.5 \, s) v = 10 \, m/s - 3.75 \, m/s v = 6.25 \, m/s The velocity after 1.5 seconds is 6.25 m/s. ii) The distance travelled in 1.5 seconds. Step 1: Identify the given values and the formula. Initial velocity u = 10 \, m/s. Acceleration a = -2.5 \, m/s^2. Time t = 1.5 \, s. We use the kinematic equation: s = ut + (1)/(2)at^2. Step 2: Substitute the values into the formula and calculate the distance. s = (10 \, m/s)(1.5 \, s) + (1)/(2)(-2.5 \, m/s^2)(1.5 \, s)^2 s = 15 \, m + (1)/(2)(-2.5 \, m/s^2)(2.25 \, s^2) s = 15 \, m - (1.25 \, m/s^2)(2.25 \, s^2) s = 15 \, m - 2.8125 \, m s = 12.1875 \, m The distance travelled in 1.5 seconds is 12.19 m. iii) The time taken by the car to stop. Step 1: Identify the given values and the formula. Initial velocity u = 10 \, m/s. Final velocity v = 0 \, m/s (when the car stops). Acceleration a = -2.5 \, m/s^2. We use the kinematic equation: v = u + at. Step 2: Substitute the values into the formula and solve for time t. 0 \, m/s = 10 \, m/s + (-2.5 \, m/s^2)t -10 \, m/s = (-2.5 \, m/s^2)t t = -10 \, m/s-2.5 \, m/s^2 t = 4 \, s The time taken by the car to stop is 4 s. iv) Sketch the velocity-time graph for the motion of the car up to the time the car stopped. The velocity-time graph will be a straight line. • The initial velocity at t=0 \, s is 10 \, m/s. • The final velocity at t=4 \, s (from part iii) is 0 \, m/s. • The graph starts at (0, 10) and ends at (4, 0). • The slope of the line represents the acceleration, which is -2.5 \, m/s^2. v) From the graph determine the distance covered by the car before it stopped. Step 1: The distance covered is the area under the velocity-time graph. The graph forms a triangle with the time axis. Base of the triangle = time taken to stop = 4 \, s (from part iii). Height of the triangle = initial velocity = 10 \, m/s. Step 2: Calculate the area of the triangle. Distance = (1)/(2) × base × height Distance = (1)/(2) × (4 \, s) × (10 \, m/s) Distance = 20 \, m The distance covered by the car before it stopped is 20 m. 15. a) i) Determine the acceleration of gravity on the planet. Step 1: Identify the given values and the formula for the frequency of a simple pendulum. Length of pendulum L = 0.5 \, m. Frequency f = 0.29 \, Hz. The mass of the body (50 \, g) is not needed for this calculation. The formula for the frequency of a simple pendulum is: f = (1)/(2) sqrt((g)/(L)). Step 2: Rearrange the formula to solve for g. 2 f = sqrt((g)/(L)) (2 f)^2 = (g)/(L) g = L(2 f)^2 Step 3: Substitute the values and calculate g. g = (0.5 \, m)(2 × 0.29 \, Hz)^2 g = (0.5 \, m)(1.8221237 \, s^-1)^2 g = (0.5 \, m)(3.32014 \, s^-2) g = 1.66007 \, m/s^2 Rounding to two significant figures: g ≈ 1.7 \, m/s^2 The acceleration of gravity on the planet is 1.7 m/s^2. Send me the next one 📸