Let’s break down the two questions in steps, with explanations: --- ## **Step 1: Analyzing theeleration Graph (d)** - **To sketch the acceleration graph**, we need to know how the cart’s velocity changes with time. - Unfortunately, the velocity-time graph is not provided here. However, based on the context of the student discussion, it seems there is a moment (at t = 6 s) where the cart’s motion is important—possibly a turning point like stopping or changing direction. - If the **velocity changes direction at t = 6 s**, then at that instant, the cart's velocity is zero, and it may involve a change in acceleration (possibly a spike if instantaneous, or a step if constant acceleration switches direction). **Explanation:** If the cart moves with **constant acceleration**, the acceleration graph would show a horizontal line (above or below zero, depending on direction). If the acceleration changes at t = 6 s, the graph would show a step up or down at that point. --- ## **Step 2: Analyzing the Students' Reasoning (e)** Let's break down each student's statement: ### **i. Student 1's Reasoning** - **Student 1 says:** At t = 6 s, the direction of the cart changes. - This would only be correct if the velocity crosses zero at t = 6 s (i.e., the cart stops and reverses direction). - If the velocity is still positive at t = 6 s (as Student 2 says), then Student 1 is **incorrect** about the cart changing direction at that time. **Explanation:** Student 1 is only correct **if** the velocity crosses zero at t = 6 s. If not, their reasoning is incorrect. --- ### **ii. Student 2's Reasoning** - **Student 2 says:** The velocity is still positive at 6 seconds, so it must be continuing in the same direction. - This is correct. As long as the velocity is positive (and not zero or negative), the cart is moving in the same direction. **Explanation:** Student 2 correctly uses the velocity value at t = 6 s to deduce that the direction has not changed. --- ## **Final Answers** ### d) **Sketch of Acceleration vs. Time** - **If acceleration is constant:** Draw a horizontal line (e.g., above or below zero). - **If acceleration changes at t = 6 s:** Draw a step in the graph at t = 6 s (e.g., from positive to negative or vice versa). - **Without the velocity-time graph, the best answer is to note that any change in acceleration at t = 6 s should be reflected as a step or spike in the acceleration graph.** --- ### e) **Student Reasoning** i. **Student 1 is only correct if the velocity crosses zero at t = 6 s.** If the velocity is still positive, Student 1's reasoning is incorrect. ii. **Student 2 is correct:** If the velocity is still positive at t = 6 s, the cart is still moving in the same direction. --- **Final summary:** - **d)** The acceleration graph should reflect any changes at t = 6 s (step or spike). - **e.i)** Student 1 is only correct if velocity = at t = 6 s. - **e.ii)** Student 2 is correct: positive velocity means same direction.