When the Derivative Misleads: A Piecewise Motion Problem
A particle moves along a line with position s(t) in metres given piecewise by s(t) = t^2 for 0 <= t <= 2 and s(t) = 4 + 3(t - 2) for 2 < t <= 5, where t is in seconds. A student computes the derivative at t = 2 by differentiating the first piece, getting s'(2) = 4, and concludes the particle's velocity at t = 2 is 4 m/s. Another student differentiates the second piece and gets velocity 3 m/s at t = 2. The teacher notes the position graph has a corner at t = 2. The class must decide what the velocity actually is at t = 2 and how to describe the motion over the whole interval.
1. Determine whether s is differentiable at t = 2, and state the velocity there.
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Check one-sided derivatives. From the left, s'(t) = 2t, so the left derivative at t = 2 is 4. From the right, s'(t) = 3, so the right derivative is 3. Since 4 is not equal to 3, the derivative does not exist at t = 2; the graph has a corner. Velocity is the derivative of position, so velocity is undefined at the instant t = 2. The first student's answer of 4 is the left-hand velocity; the second's 3 is the right-hand velocity. Neither is 'the' velocity. The judgement is that piecewise functions require checking one-sided limits before quoting a derivative, and that 'undefined' is a legitimate, markable answer.
2. Describe the particle's speed and direction over 0 <= t <= 5, including what happens at t = 2.
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For 0 <= t < 2: s'(t) = 2t >= 0, so the particle moves in the positive direction, speeding up from 0 to a left-limit velocity of 4 m/s. At t = 2 the velocity is undefined (a sudden change, an impulse-like corner). For 2 < t <= 5: s'(t) = 3 > 0, so it continues in the positive direction at a constant 3 m/s. The particle never reverses direction; it only changes speed abruptly at t = 2. Position is continuous at t = 2: both pieces give s(2) = 4, so the particle does not jump, only its velocity does. Distinguishing continuity of position from differentiability of position is the core judgement here.
3. Find the total distance travelled over 0 <= t <= 5 and compare it with the displacement.
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Because velocity is non-negative throughout (2t >= 0 on [0,2), 3 > 0 on (2,5]), the particle never reverses, so distance travelled equals displacement. Displacement = s(5) - s(0). s(0) = 0. s(5) = 4 + 3(5 - 2) = 4 + 9 = 13. So displacement = 13 m and total distance = 13 m. Check continuity at t = 2: s(2) = 4 from both pieces, so no gap. The judgement is recognizing that distance equals displacement only when there is no sign change in velocity; here the corner changes speed but not direction. A student who assumed the corner implied a reversal would wrongly add segments and overcount.