AS 91579 · 2025 paper · substitution-type, constants, kinematics
Question 2(d) Merit
Practises Differential equations
A power-of-a-linear integral, a rate of change integrated back to its function, a constant found from a definite integral, a kinematics chain (acceleration → velocity → displacement), and an odd-power trig area.
A particle’s acceleration can be modelled by a(t) = 2.4e−0.3t − 1, where a(t) is in m s⁻² and t is the time in seconds from the start of timing. Initially, at a fixed point P, the particle had a velocity of 6 m s⁻¹.
How far from the point P is the particle 3 seconds after timing started?
| a(t) | = | 2.4e−0.3t − 1 m s⁻² |
|---|---|---|
| v(0) | = | 6 m s⁻¹ |
| t | = | 3 s |
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Velocity is the integral of acceleration. Integrate a(t).
v = int (2.4e−0.3t − 1)dt = (2.4/(−0.3))e−0.3t − t + c = −8e−0.3t − t + c -
2 Use v = 6 when t = 0.
6 = −8e0 − 0 + c = −8 + c, so c = 14 and
v = −8e−0.3t − t + 14 -
3 Displacement is the integral of velocity. Integrate again, with a new constant.
s = ((−8)/(−0.3))e−0.3t − t2/2 + 14t + d = (80/3)e−0.3t − t2/2 + 14t + d -
4 The particle starts at P, so s = 0 when t = 0. Does that make d zero?
No. At t = 0 the exponential term is (80/3)e0 = 80/3, so 0 = 80/3 + d and d = −80/3. -
5 Now put t = 3 in.
s = (80/3)e−0.9 − 9/2 + 42 − 80/3 = 21.675 m
What the marker wanted
- Achieved The velocity function with its constant found: v = −8e−0.3t − t + 14.
- Merit The distance, 21.675 m, with the integration shown.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Taking the second constant as zero because the particle starts at P. NZQA’s 2025 report says a large number of students did this. s = 0 at t = 0 makes the whole expression zero, and its exponential term is 80/3 there, not 0.How did it go?
Got it moves it further out. Review brings it back tomorrow.