Bike Race

We let $A(d)$ and $C(d)$ be the time it takes Alice and Carla to travel a distance $d$ , respectively. We need to find the maximum value of $d$ such that $C(d)<A(d$ .

To find $A(d)$ , we first need to find Alice's acceleration. Alice is riding down an inclined plane, meaning that the force acting on her is the parallel force due to gravity (I wish you could post photos to comments it would make explaining this a lot easier). The parallel force is $mg\sin(\theta)$ , where $\theta$ is the angle of the incline. Now this is the force that dictates how fast she travels parallel to the hill. We need to find how fast she travels parallel to the ground. This formula is simply $(mg\sin(\theta))(\cos(\theta))$ . Finally, using our formula $F=ma$ , solving for $$ and plugging in $45^o$ as our angle:

$a=\frac{F}{m}=(10m/s^2)(\frac{\sqrt{2}}{2})^2=5m/s^2$

Her formula $A(d)$ , derived from $d=\frac{1}{2}a(A(d))^2$ , must be:

$A(d)=\sqrt{\frac{2d}{5}}$

Now to find $C(d)$ . We know that Carla starts off at $0m/s$ , accelerates at $10m/s^2$ , then stops after $2$ seconds at $20m/s^2$ . Her distance traveled after $2$ seconds then must be $d=\frac{1}{2}at^2=\frac{1}{2}(10m/s^2)(2s)^2=20m$ . After two seconds she rides at a constant rate of $20m/s$ . Her distance traveled during this time can be written as $d=vt=(20m/s)(t-2)$ . Summing the two values of $d$ together, replacing $t$ with $C(d)$ and solving for $C(d)$ we get:

$C(d)=\frac{d}{20}+1$

Finally, let's plug these equations into our inequality $C(d)<A(d)$ and solve for d!

$\frac{d}{20}+1<\sqrt{\frac{2d}{5}}$

$d<20(3+2\sqrt{2})=116.569...$

Therefore our final answer must be $\boxed{116}$ .

Carla's motion is described as $C(t) = \left\{\begin{array}{ll} 5t^2 & t \leq 2 \\ 20 + 20 (t-2) & t \geq 2 \end{array}\right.$ Alice's motion is simply $A(t) = 2\tfrac12 t^2.$ (The $45^\circ$ angle causes Alice's acceleration to be $\tfrac12\sqrt 2 g$ , and to find the x-component of this acceleration we must once again multiply by $\tfrac12\sqrt2$ , so that the horziontal acceleration is $\tfrac12g$ .) Clearly Carla will be ahead of Alice during the first two seconds; therefore we equate $20 + 20(t-2) = 2\tfrac12 t^2.$ This translates to $t^2 - 8t + 8 = 0 \therefore t = 4+\sqrt 8 = 6.83\ \text{s};$ substitute this in both equations to find $A(t) = C(t) = 116.6\ \text{m}.$ Thus the answer is 116 meters.