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 are the solutions for the questions. Multiple Choice Question: The graph shows a linear relationship between velocity (V) and time (t), which represents motion with constant acceleration. The general equation for velocity as a function of time under constant acceleration is v = u + at, where u is the initial velocity and a is the constant acceleration. Option B: v - u = at. This can be rearranged to v = u + at. Option D: v = u + at. Both options B and D represent the correct kinematic equation for the velocity-time graph shown. Since option B is circled, we will confirm it. The correct option is B. --- Question 6: The graph shows the velocity-time relationship for an object projected upwards, reaching a maximum height, and returning to its point of projection. I) How long does it take to reach maximum height? Step 1: Identify the condition for maximum height. The object reaches its maximum height when its vertical velocity momentarily becomes zero before it starts falling back down. Step 2: Read the time from the graph. From the graph, the velocity V is 0 \, m/s at t = 4 \, s. It takes 4 seconds to reach maximum height. II) What is the total time of flight? Step 1: Relate time to maximum height and total time of flight. For an object projected upwards and returning to its initial point (assuming no air resistance), the time taken to reach the maximum height is equal to the time taken to fall back from the maximum height to the initial point. Step 2: Calculate the total time of flight. Time to reach maximum height = 4 \, s. Total time of flight = 2 × (time to reach maximum height) Total time of flight = 2 × 4 \, s = 8 \, s. The total time of flight is 8 seconds. III) Which time interval shows the object returning? Step 1: Identify the condition for the object returning. The object is returning when its velocity is negative, indicating downward motion. Step 2: Determine the time interval from the graph. From the graph, the velocity becomes negative after t = 4 \, s. The object returns to its point of projection at t = 8 \, s (as calculated in part II). The object is returning in the time interval from 4 \, s < t 8 \, s. IV) What will be the maximum velocity at 4 seconds? Step 1: Identify the velocity at maximum height. At t = 4 \, s, the object reaches its maximum height. At this point, its vertical velocity is momentarily zero. The velocity at 4 seconds is 0 \, m/s. V) Draw a displacement-time graph. To describe the displacement-time graph, we first need to determine the initial velocity and acceleration from the given velocity-time graph. Step 1: Determine the acceleration. The slope of the velocity-time graph represents the acceleration. From the graph, the velocity is 0 \, m/s at t = 4 \, s. Let's assume the initial velocity u at t=0 is 40 \, m/s (consistent with the graph's appearance if g ≈ 10 \, m/s^2). a = ( V)/( t) = 0 \, m/s - 40 \, m/s4 \, s - 0 \, s = -40 \, m/s4 \, s = -10 \, m/s^2 Step 2: Calculate the maximum displacement (maximum height). The displacement is the area under the velocity-time graph. The maximum height is reached at t=4 \, s. s_max = Area of triangle from t=0 to t=4 \, s s_max = (1)/(2) × base × height = (1)/(2) × 4 \, s × 40 \, m/s = 80 \, m Step 3: Describe the displacement-time graph. The displacement-time graph for an object under constant acceleration (like gravity) is a parabola. • The graph starts at (0,0) (initial displacement is zero). • It increases parabolically, reaching a maximum displacement of 80 \, m at t=4 \, s. • It then decreases parabolically, returning to zero displacement at t=8 \, s (the total time of flight). • The overall shape is a downward-opening parabola. 3 done, 2 left today. You're making progress.