Quadratic question problem

Question

I have been asked to look at a yr 12 question about a space ships trajectory modelled by a quadratic equation. But my solution to the question has a negative minimum for distance at a time of T = 13 minutes. I was wondering if anyone can confirm that the question is suspicious or show how to work out a correct answer. I believe that this question has been used in exams before so I am puzzled as to why I am getting a negative value for distance at the minimum. Any help would be appreciated.

An on line calculator provided a quadratic solution of $$y = 3x^2 - 78x + 500$$ Whose minimum is at the discriminant $\frac{-b}{2a} = 13$ And so the minimum height is $$3 \times 13^2-78 \times 13+500 = -7 m$$

See image for the question.

Yr 12 Math Exam Question

Regards Michael Mckeon

Just a nitpick but that should be $-7 \mathrm{km}$, i.e. $7$ kilometres (not metres) below the surface. — Deepak, Jun 18 '15 at 00:29
The distance reported reflects that how far away is the rocket from the point of contact. In the calculation, the exact value needed is the displacement (a vector quantity) instead. It is reported in the opposite direction of travel. Suggestion: $y = –(3x^2 - 78x + 500)$. — Mick, Jun 20 '15 at 04:48

Deepak · Answer 1 · 2015-06-18T00:35:37.797

I got the same quadratic equation and solution you did. Brief summary of method (since this is an answer, not just a comment):

Let $d(t) = at^2 + bt + c$

With the three given points on the trajectory, we have the simultaneous equations:

$425 = a + b + c$

$356 = 4a + 2b + c$

$293 = 9a + 3b + c$

Solving, $a = 3, b = -78, c = 500$

So $d(t) = 3t^2 - 78t + 500$

The derivative wrt $t$ is:

$d'(t) = 6t - 78$

Setting that to zero:

$6T = 78 \implies T = 13$

and $d(T) = 3(13)^2 - 78(13) + 500 = -7$

I found the minimum using basic calculus (another approach) and got the same answer you got (i.e. the ship would end up buried $7 \mathrm{km}$ beneath the Martian surface if you completely neglected the fact that it would wreck against the surface). The question is definitely "broken", about as broken as the poor astronaut's ship is going to be if he tries to keep going for $13$ seconds after switching on his retro rockets.

It's easy to solve for when he collides with the surface. Set $d(\tau) = 0$, giving:

$3\tau^2 - 78\tau + 500 = 0$

Solving, we get:

$\tau \approx 11.47$ or $\tau \approx 14.53$.

Reject the greater solution, so the ship collides $11.47\mathrm{s}$ after the retro rockets are activated.

Just for fun, it is easy to show that the ship, just before impact, would be travelling at a bit over $9 \mathrm{km/s}$, which is over $32,000 \mathrm{km/h}$ or Mach $27$. The remains are gonna be pretty mangled. — Deepak, Jun 18 '15 at 01:10
Thanks for all your responses, these have been very helpful. — Michael T Mckeon, Jun 21 '15 at 05:29

score 0 · Answer 2 · answered Jun 20 '15 at 05:25

0

This is only a suggestion on how to treat this question.

enter image description here

If the maximum displacement was found to be s, then d - s is the (shortest) distance from the target.

answered Jun 20 '15 at 05:25

Mick

17,141

Thanks mick, that's an interesting interpretation, but I suspect that the question is at least ambiguous. – Michael T Mckeon Jun 21 '15 at 05:28
@MichaelTMckeon Does this book have a name? Does this question have a suggested answer? – Mick Jun 23 '15 at 17:49
No its not a book, the question was used a couple of times in an exam. Thanks again for your help. – Michael T Mckeon Jun 25 '15 at 13:37

Quadratic question problem

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