91Ó°ÊÓ

Question: Using the procedure outlined above, sketch a velocity versus time graph for a car starting from rest, accelerating for 5 seconds, maintaining constant velocity for 3 seconds, and then decelerating for 6 seconds back to a complete stop. Determine the car's acceleration during each interval of motion. Answer: To sketch the graph, identify the key moments in the car's motion: 1. Starts from rest (initial velocity \(v_0 = 0 \frac{m}{s}\)). 2. Accelerates for 5 seconds (assume final velocity \(v_1 = 10 \frac{m}{s}\)). 3. Maintains constant velocity for 3 seconds (final velocity \(v_2 = 10 \frac{m}{s}\)). 4. Decelerates for 6 seconds back to a complete stop (final velocity \(v_3 = 0 \frac{m}{s}\)). Now, plot these points on the velocity versus time graph: 1. \((0, 0)\) represents starting from rest. 2. \((5, 10)\) represents the end of acceleration period. 3. \((8, 10)\) represents the end of the constant velocity period. 4. \((14, 0)\) represents the end of deceleration period and a complete stop. Connect the points with line segments: 1. Connect \((0, 0)\) to \((5, 10)\) with a straight line, representing the acceleration phase. 2. Draw a horizontal line from \((5, 10)\) to \((8, 10)\), representing constant velocity. 3. Connect \((8, 10)\) to \((14, 0)\) with a straight line, representing deceleration back to a complete stop. Now, analyze the graph for acceleration: 1. During the acceleration phase, the slope or acceleration is \(\frac{\Delta v}{\Delta t} = \frac{10 - 0}{5 - 0} = \frac{10}{5} = 2 \frac{m}{s^2}\). 2. During the constant velocity phase, the slope is zero, so the acceleration is \(0 \frac{m}{s^2}\). 3. During the deceleration phase, the slope or deceleration is \(\frac{\Delta v}{\Delta t} = \frac{0 - 10}{14 - 8} = \frac{-10}{6} = -\frac{5}{3} \frac{m}{s^2}\). Thus, the car's acceleration during each interval of motion is as follows: - Acceleration phase: \(a = 2 \frac{m}{s^2}\) - Constant velocity phase: \(a = 0 \frac{m}{s^2}\) - Deceleration phase: \(a = -\frac{5}{3} \frac{m}{s^2}\)