Toy Car Speed On Circular Track: A Trigonometry Problem

Hey guys! Let's dive into a fun math problem involving a toy car, a circular track, and some trigonometry. We're going to explore how a student uses the equation $\tan \theta = \frac{s^2}{49}$ to figure out the speed, s, of the car. This equation relates the speed of the toy car to the angle of incline, $\theta$, of the track. We also know that $\sin \theta = \frac{1}{2}$. Sounds interesting, right? Let's break it down step by step.

Understanding the Problem Setup

First, let's make sure we all understand what's going on in this problem. We've got a toy car zooming around a circular track. This track isn't flat; it's inclined at an angle, which we're calling $\theta$. Think of it like a mini-racetrack that's tilted slightly. The car's speed, s, is measured in feet per second. The equation $\tan \theta = \frac{s^2}{49}$ is the key to linking the speed to the angle of the incline. This formula comes from physics, where the centripetal force required to keep the car moving in a circle is related to the angle of the incline and the car's speed. The number 49 likely incorporates the radius of the circular path and the acceleration due to gravity. We are also given that $\sin \theta = \frac{1}{2}$. This piece of information is crucial because it allows us to find the value of $\theta$ and subsequently $\tan \theta$, which we can then use in our main equation. So, the big question is: How do we use all this information to find the speed, s, of the toy car? That's what we're going to figure out together. Understanding the relationship between the angle, speed, and the physics of circular motion is essential for solving this problem. Let's move on to the next step: figuring out the angle $\theta$.

Finding the Angle θ

Now, let's figure out the value of the angle $\theta$. We're given that $\sin \theta = \frac{1}{2}$. Remember your trigonometry basics? We need to find the angle whose sine is one-half. Think about the unit circle or your special right triangles. The sine function corresponds to the y-coordinate on the unit circle, and we know that $\sin 30^\circ = \sin \frac{\pi}{6} = \frac{1}{2}$. So, the angle $\theta$ is 30 degrees, or $\frac{\pi}{6}$ radians. This is a crucial step because now that we know $\theta$, we can find the value of $\tan \theta$. We're one step closer to finding the speed of the toy car! Remember, the sine, cosine, and tangent functions are all related, and knowing one can help you find the others. In this case, knowing the sine allowed us to determine the angle, which is vital for using the tangent equation. It's like unlocking a secret code to the problem. With the angle $\theta$ in hand, we're ready to tackle the next challenge: calculating $\tan \theta$. This will directly link us to the car's speed through the given equation. Let’s move on to figuring out what $\tan 30^\circ$ is.

Calculating tan θ

Alright, we've found that $\theta = 30^\circ$ (or $\frac{\pi}{6}$ radians). Now we need to calculate $\tan \theta$, which means finding $\tan 30^\circ$. Again, think back to your trig basics – those special right triangles are super helpful here! Consider a 30-60-90 triangle. The sides are in the ratio 1:$\sqrt{3}$:2, where 1 is opposite the 30-degree angle, $\sqrt{3}$ is opposite the 60-degree angle, and 2 is the hypotenuse. Tangent is defined as the opposite side divided by the adjacent side. So, $\tan 30^\circ = \frac{1}{\sqrt{3}}$. To rationalize the denominator, we multiply both the numerator and the denominator by $\sqrt{3}$, which gives us $\tan 30^\circ = \frac{\sqrt{3}}{3}$. This is another key piece of the puzzle. We now have a numerical value for $\tan \theta$, which we can plug into our original equation. Remember, the goal is to find s, the speed of the car. We're methodically working our way through the problem, using trigonometry to connect the angle of incline to the car's velocity. Now that we know $\tan \theta$, we're ready to use the equation $\tan \theta = \frac{s^2}{49}$ to solve for s. Let's do it!

Solving for the Speed, s

Okay, this is where the magic happens! We have the equation $\tan \theta = \frac{s^2}{49}$, and we know that $\tan \theta = \frac{\sqrt{3}}{3}$. Let's substitute that value into the equation: $\frac{\sqrt{3}}{3} = \frac{s^2}{49}$. Now, we need to solve for s. First, we'll multiply both sides of the equation by 49 to get rid of the fraction on the right side: $s^2 = 49 \cdot \frac{\sqrt{3}}{3}$. This simplifies to $s^2 = \frac{49\sqrt{3}}{3}$. To find s, we need to take the square root of both sides: $s = \sqrt{\frac{49\sqrt{3}}{3}}$. We can simplify this a bit further. The square root of 49 is 7, so we have $s = 7\sqrt{\frac{\sqrt{3}}{3}}$. To get a more user-friendly answer, we can approximate the square root. $\sqrt{3}$ is approximately 1.732. So, we have: $s = 7\sqrt{\frac{1.732}{3}} \approx 7\sqrt{0.577} \approx 7 \cdot 0.76 \approx 5.32$. Therefore, the speed s is approximately 5.32 feet per second. We've successfully used trigonometry and algebra to solve for the speed of the toy car! It's awesome how math can describe real-world situations, isn't it? Let’s recap the entire process to make sure we’ve nailed it.

Recap and Final Thoughts

Alright, let's take a step back and recap what we've done. We started with a cool problem: a toy car racing around an inclined circular track. We were given the equation $\tan \theta = \frac{s^2}{49}$, which links the car's speed s to the angle of incline $\theta$. We also knew that $\sin \theta = \frac{1}{2}$. Our mission was to find the speed s. First, we used the information $\sin \theta = \frac{1}{2}$ to determine that $\theta = 30^\circ$. Then, we calculated $\tan 30^\circ$, which is $\frac{\sqrt{3}}{3}$. Next, we plugged this value into our main equation: $\frac{\sqrt{3}}{3} = \frac{s^2}{49}$. We solved for $s^2$ and then took the square root to find s. We ended up with $s \approx 5.32$ feet per second. So, the toy car is cruising at about 5.32 feet per second around the track. This problem is a great example of how trigonometry and algebra can be used together to solve real-world problems. We used the sine function to find an angle, the tangent function to relate that angle to the car's speed, and then algebraic manipulation to isolate and solve for the unknown speed. Math is pretty awesome, huh? I hope you guys found this breakdown helpful and maybe even a little fun. Keep practicing these types of problems, and you'll become a math whiz in no time! Remember, the key is to break down the problem into smaller, manageable steps and to use the information you have to find what you need. Great job working through this with me!