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Q2.18: NCERT Physics - Part 1 MOTION IN A STRAIGHT LINE 2.18 Solution: Figure 2.15 gives a speed-time graph of a particle in motion…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often incorrectly assume that the acceleration at a specific point (like B or C) is zero because the velocity is momentarily constant or at a peak, failing to recognize that instantaneous acceleration is determined by the slope of the tangent to the velocity-time graph at that exact point, not the value of the velocity itself.
Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?
Step-by-Step Verified Solution
**Answer:** Average acceleration is greatest in interval 2. Average speed is greatest in interval 3. In all intervals v > 0. In interval 1: a > 0; interval 2: a < 0; interval 3: a > 0. Accelerations at A, B, C, D: a_A > 0, a_B < 0, a_C = 0, a_D > 0.
**Step-by-Step Explanation:**
Average acceleration = _v/_t. Interval 2 has the largest |_v|. Average speed = area under v-t graph / time. Interval 3 has the largest area. Signs of v are positive (constant direction). Signs of a from slope: interval 1 positive, interval 2 negative, interval 3 positive. At A slope positive (a > 0), at B slope negative (a < 0), at C slope zero (a = 0), at D slope positive (a > 0).
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