Maths · Calculus · chapter 12 of 14
Class 11 Maths Chapter 12: Limits and Derivatives case study question with answers
One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A scooter in Kochi enters a straight stretch of road at time t = 0. A roadside sensor records that, t seconds later, it has covered s(t) = t2 + 4t metres of the stretch. The average velocity over an interval is the distance covered in it divided by the time taken. The velocity at an instant is the limit of the average velocities over shorter and shorter intervals starting at that instant, which is the derivative of s.
Part 1 · 1 mark
The average velocity of the scooter between t = 2 s and t = 4 s is
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Answer: (b) 10 m/s
s(4) = 32 m and s(2) = 12 m, so the average velocity is 20/2 = 10 m/s.
Part 2 · 1 mark
The velocity of the scooter at t = 2 s is
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Answer: (c) 8 m/s
s′(t) = 2t + 4, so s′(2) = 8 m/s.
Part 3 · 2 marks
Using the first principle, find the velocity of the scooter at time t, and the time at which it is moving at 14 m/s.
Answer and marking scheme
s(t + h) − s(t) = 2th + h2 + 4h, so (s(t + h) − s(t))/h = 2t + h + 4, which tends to 2t + 4 as h → 0. Setting 2t + 4 = 14 gives t = 5 s.
- v(t) = 2t + 4 from the first principle (1)
- t = 5 s (1)
Or
Find the average velocity of the scooter over the interval from t = 3 s to t = (3 + h) s, and hence its velocity at t = 3 s.
Answer and marking scheme
s(3 + h) − s(3) = (9 + 6h + h2 + 12 + 4h) − 21 = 10h + h2, so the average velocity is 10 + h m/s. As h → 0 it tends to 10, so the velocity at t = 3 s is 10 m/s.
- average velocity 10 + h (1)
- limit 10 m/s as h → 0 (1)
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