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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Here's the solution to question 9: The car's journey can be broken down into three phases: Phase 1: Acceleration Starts from rest (0 m/s). Reaches 40 m/s. Time taken: 1 min 40 s = 60 s + 40 s = 100 s. End of phase 1: (100 s, 40 m/s). Phase 2: Constant speed Speed: 40 m/s. Time taken: 5 min = 5 × 60 s = 300 s. This phase starts at 100 s and ends at 100 s + 300 s = 400 s. End of phase 2: (400 s, 40 m/s). Phase 3: Deceleration Starts at 40 m/s. Comes to a halt (0 m/s). Time taken: 2 min = 2 × 60 s = 120 s. This phase starts at 400 s and ends at 400 s + 120 s = 520 s. End of phase 3: (520 s, 0 m/s). a) Draw a velocity-time graph to represent the information above. To draw the graph: 1. Draw a horizontal axis for time (in seconds) and a vertical axis for velocity (in m/s). 2. Plot the following points: (0, 0) (start from rest) (100, 40) (end of acceleration) (400, 40) (end of constant speed) (520, 0) (car comes to a halt) 3. Connect (0,0) to (100,40) with a straight line (acceleration). 4. Connect (100,40) to (400,40) with a horizontal line (constant velocity). 5. Connect (400,40) to (520,0) with a straight line (deceleration). The resulting graph will be a trapezium. b) Use your graph to find: (i) The initial acceleration. The initial acceleration is the gradient of the first segment of the graph (from t=0 s to t=100 s). Acceleration = Change in velocityChange in time Acceleration = 40 m/s - 0 m/s100 s - 0 s Acceleration = (40)/(100) m/s^2 Acceleration = 0.4 m/s^2 The initial acceleration is 0.4 m/s^2. (ii) The deceleration when the car is brought to a halt. The deceleration is the magnitude of the gradient of the last segment of the graph (from t=400 s to t=520 s). Acceleration = Change in velocityChange in time Acceleration = 0 m/s - 40 m/s520 s - 400 s Acceleration = -40 m/s120 s Acceleration = -(1)/(3) m/s^2 The deceleration is the positive value of this acceleration. Deceleration = (1)/(3) m/s^2 ≈ 0.333 m/s^2 The deceleration is 0.333 m/s^2 (to 3 significant figures). (iii) The distance traveled in km. The total distance traveled is the area under the entire velocity-time graph (a trapezium). The formula for the area of a trapezium is (1)/(2) × (sum of parallel sides) × height. • Height of the trapezium (maximum velocity) = 40 m/s. • Length of the top parallel side (duration of constant velocity) = 400 s - 100 s = 300 s. • Length of the bottom parallel side (total time) = 520 s. Total distance = (1)/(2) × (300 s + 520 s) × 40 m/s Total distance = (1)/(2) × 820 s × 40 m/s Total distance = 410 s × 40 m/s Total distance = 16400 m To convert meters to kilometers: Distance in km = 16400 m1000 m/km Distance in km = 16.4 km The total distance traveled is 16.4 km. Last free one today — make it count tomorrow, or type /upgrade for unlimited.