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.
Regards Michael Mckeon
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.