# Mechanics - Kinematics in two dimensions, how to find the time given a quadratic?

I have this information: $$u = 2i + 3j$$ $$r0(\text{initial position}) = 40i + 20j$$ $$r = 52i + 128j$$ $$a = -0.06i -0.04j$$

I need to find the $t(\text{time})$ at this point. I can use the equation $r = ut + \frac{1}{2}at^2 + r0$.

Substituting in the vectors I have I get:

$$52i + 128j = 2ti + 3tj + \frac{1}{2}\left(-0.06t^2\right)i + \frac{1}{2}\left(-0.04t^2\right)j + 40i + 20j$$

Solving down, I get:

$$12i + 108j = \left(2t - 0.03t^2\right)i + \left(3t - 0.02t^2\right)j$$

From here, I tried just solving as a quadratic with just the $i$ components, and get $t = 60$(correct answer) and $t = 6.66666$. If I solve using the $j$ component, I get $t = 90$ and $t = 60$. I don't understand how to find out which is the correct answer to use in a situation without the answers to hand.

• You need a value of $t$ which works for both components simultaneously. Hence the only possibility is _____. – Macavity Feb 19 '15 at 12:01
• Oh blimey... I see now, cannot believe I missed that. Thanks very much! – Ed Prince Feb 19 '15 at 12:02