AS 91579 · 2023 paper · Simpson’s rule, a trig area, a factorised differential equation
Question 3(d) Merit
Practises How to choose u
Opens with Simpson’s Rule on five measured offsets from a garden path, then a root quotient, a product-of-sines area, an exponential-quotient acceleration integrated to a velocity, and a differential equation that has to be factorised before it will separate.
An object’s acceleration can be modelled by a = e2t/(4e2t − 3), where t ≥ 0, a is in m s⁻² and t is the time in seconds from the start of timing. At t = 0 the object had a velocity of 5 m s⁻¹.
Find the object’s velocity when t = 4 seconds.
| a | = | e2t/(4e2t − 3) m s⁻² |
|---|---|---|
| v(0) | = | 5 m s⁻¹ |
| t | = | 4 s |
Work it out
Try it on paper first. Stuck? Take one hint at a time.
-
1 Velocity is the integral of acceleration. How does the top compare with the derivative of the bottom?
(d/dt)(4e2t − 3) = 8e2t: the top, times 8. -
2 So what does it integrate to?
v = (1/8)int (8e2t/(4e2t − 3)) dt = (1/8) ln(4e2t − 3) + c. For t ≥ 0 the bracket is at least 1, so no modulus is needed. -
3 Use v = 5 when t = 0.
4e0 − 3 = 1 and ln 1 = 0, so c = 5. -
4 Put t = 4 in.
v = (1/8) ln(4e8 − 3) + 5 = 6.1733 m s⁻¹
What the marker wanted
- Achieved The equation for v, (1/8) ln(4e2t − 3) + c, from correct integration.
- Merit The velocity, about 6.17 m s⁻¹, from correct integration.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.