The diameter of a car's tire is 60cm. While the car is being driven, the tire picks up a nail. At what times in two complete cycles is the nail at a height of 1cm above the ground.
So far, I have calculated the circumference of the wheel (188.5) and I do not know the next step.
This is pretty "advanced" stuff
Your nail traces out a cycloid, the
equation of that path can be written as
parametric equations
x=r(t−sint)
y=r(1−cost) , where t is the angle of rotation, and r is the radius
What level of math is this ?
http://mathworld.wolfram.com/Cycloid.html
Grade 11 Unversity Functions.
The question was part of the final "Thinking" category for our periodic functions test.
To find the times when the nail is at a height of 1cm above the ground, we can use the fact that the height of the nail will repeat every time the tire completes one full rotation.
Since the diameter of the tire is 60cm, the radius (r) would be half of that, which is 30cm.
The circumference of a circle is calculated using the formula: C = 2πr.
So, the circumference of the tire is:
C = 2π(30)
C = 60π cm
Now, we need to find the time it takes for the tire to complete one full rotation. Let's assume the car is traveling at a constant speed. If we know the speed of the car, we can calculate the time it takes to complete one revolution.
Let's say the speed of the car is v meters per second.
The formula to find the time (t) it takes for an object to travel a certain distance (d) at a constant speed (v) is: t = d/v.
In this case, the distance is the circumference of the tire and the speed is the distance traveled per unit time.
So, the time it takes to complete one full rotation is:
t = (60π cm)/(v cm/s)
t = (60π)/v s
Now, to find the times when the nail is at a height of 1cm above the ground, we need to find the values of t when the height repeats.
Since the height repeats every one full rotation, the nail will be at a height of 1cm above the ground at time t = n * (60π)/v, where n is any integer.
For example, if n = 0, the nail will be at a height of 1cm above the ground at the start of the cycle. And if n = 2, the nail will be at the same height after two complete cycles.
So, to find the times in two complete cycles when the nail is at a height of 1cm above the ground, you can plug in the value of n and calculate the corresponding time using the formula t = n * (60π)/v.
Note: To calculate the exact values of t, you would need to know the speed of the car (v). Without that information, you can express the times as a multiple of (60π)/v.
To find the times when the nail on the tire is at a height of 1cm above the ground, we need to consider the relationship between the height of the nail and the position of the tire as it rotates.
Since you have already calculated the circumference of the wheel, which is 188.5 cm, we can use this value to determine how far the tire travels in one complete cycle.
To find the distance the tire travels in one rotation, we use the formula:
Distance = Circumference × Number of rotations
In this case, the number of rotations is 1, so the distance traveled in one cycle is equal to the circumference, which is 188.5 cm.
Now, to determine the times when the nail is at a height of 1cm above the ground, we need to analyze the relationship between the height of the nail and the distance traveled by the tire.
Let's assume that the nail starts at the very bottom of the tire (height = 0cm) when the car begins moving. As the car moves forward, the nail will gradually rise until it reaches the topmost point of the tire (height = diameter = 60cm), and then it will start descending again.
We want to find the times when the nail is at a height of 1cm above the ground. To do this, we need to determine how many times the nail's height goes from below 1cm to above 1cm in one complete cycle.
Since the diameter of the tire is 60cm, the highest point of the tire is 60cm above the ground. For the nail to be 1cm above the ground, it has to be at a height of 61cm.
Now, let's look at how many times the nail's height goes from below 1cm to above 1cm in one complete cycle. This happens twice: once when the nail first reaches a height of 1cm during its ascent, and again when it reaches a height of 1cm during its descent.
To calculate the times at which this occurs, we divide the total distance traveled in one complete cycle (188.5 cm) by the distance traveled between each occurrence of the nail's height crossing 1cm (which is 60cm, as it covers the distance from a height below 1cm to above 1cm).
So, the times at which the nail is at a height of 1cm above the ground in two complete cycles would be:
Time 1: when the nail reaches 1cm during its first ascent
Time 2: when the nail reaches 1cm during its first descent
Time 3: when the nail reaches 1cm during its second ascent
Time 4: when the nail reaches 1cm during its second descent
To find the exact times, you would need to measure the time it takes for the tire to complete one full rotation and divide it by four, since there are four occurrences of the nail reaching a height of 1cm.
Note: The speeds at which the car is driven would affect how long it takes for the nail to reach 1cm above the ground, but the concept of dividing the distance traveled by the distance between the occurrences remains the same.