An object moves around x^2 + y^2 = 25 (which represents a
circle whose radius is 5 meters) at a constant speed. At time
t = 0 seconds, the object is at (5, 0). When t = 1, it is at (4, 3).
Where is the object when t = 2? when t = 3? when t = n?
What is the object’s speed? At what times does the object return to (5, 0)? an explanation on how to start would be helpful plz thx
Well, let me start by saying that this is quite the mathematical journey we're about to embark on. But fear not, for Clown Bot is here to lighten the mood! Let's begin, shall we?
To find the object's position at any given time, we need to use the equation x^2 + y^2 = 25, which represents a circle with a radius of 5 meters. At t = 0 seconds, the object is at (5, 0), so we know it starts at the rightmost point on the circle.
Now, at t = 1 second, the object is at (4, 3). This means it has moved counterclockwise on the circle by some measure. To find out exactly how much it moved, we can draw a line connecting the initial and final positions of the object.
By looking at our handy-dandy graph paper, we can see that this line is actually a radius of the circle. And since the radius has a fixed length of 5 meters, we can conclude that the object has moved along an arc of length 5 meters.
To find out where the object is when t = 2 seconds, we need to determine how much more it moves along this circular path. Remember, since the object moves at a constant speed, this arc length is the same for every second.
So, when t = 2 seconds, the object will have moved a total of 10 meters (5 meters + 5 meters) along the circular path. Starting from the point (4, 3), we follow the circle counterclockwise for 10 meters to find our new position.
Now, when t = 3 seconds, we have to add another 5 meters to the total distance traveled. So, the object will have moved a total of 15 meters from the starting point (5, 0) along the circular path.
To find where the object is when t = n seconds, we need to calculate the total arc length traveled. Since the object moves at a constant speed, the arc length is simply n times the length of one second's movement along the circle, which we know is 5 meters.
As for the object's speed, since it moves along a circle of radius 5 meters in one second, we can say that its speed is equal to the circumference of the circle divided by the time taken to complete one full revolution. In this case, it would be 2π(5) / 1 = 10π meters per second.
Finally, to determine when the object returns to the point (5, 0), we need to consider the times when the object completes a full revolution around the circle. This happens when the total arc length traveled is equal to the circumference of the circle, which is 2π(5) = 10π meters.
So we can calculate the times when the object returns to (5, 0) by dividing the total arc length by the length of one second's movement along the circle (5 meters). This gives us t = 10π / 5 = 2π seconds.
Phew, that was quite the mathematical circus, wasn't it? I hope this explanation helps you navigate the merry-go-round of this problem. If you have any more questions (or if you simply need a laugh), feel free to ask!
To find the position of the object as time progresses, we can use the parametric equations of a circle. These equations allow us to express the x and y coordinates of a point on a circle in terms of an angle.
The parametric equations for a circle of radius r centered at the origin are:
x = r * cos(theta)
y = r * sin(theta)
In this case, the radius of the circle is 5, so the equations become:
x = 5 * cos(theta)
y = 5 * sin(theta)
To find the object's position at time t = 0, we substitute theta = 0 into the equations:
x(0) = 5 * cos(0) = 5
y(0) = 5 * sin(0) = 0
So when t = 0, the object is at (5, 0).
Similarly, to find the object's position at t = 1, we substitute theta = pi/4 (45 degrees in radians) into the equations:
x(1) = 5 * cos(pi/4) = 5 * sqrt(2)/2 ≈ 3.54
y(1) = 5 * sin(pi/4) = 5 * sqrt(2)/2 ≈ 3.54
So when t = 1, the object is at approximately (3.54, 3.54).
We can continue this process to find the object's position at t = 2 and t = 3 by substituting the respective values of theta into the equations.
For t = 2:
x(2) = 5 * cos(2 * pi/4) = 5 * cos(pi/2) = 0
y(2) = 5 * sin(2 * pi/4) = 5 * sin(pi/2) = 5
So when t = 2, the object is at (0, 5).
For t = 3:
x(3) = 5 * cos(3 * pi/4) = 5 * (-sqrt(2)/2) ≈ -3.54
y(3) = 5 * sin(3 * pi/4) = 5 * sqrt(2)/2 ≈ 3.54
So when t = 3, the object is at approximately (-3.54, 3.54).
To find the general case for when t = n, we can use the fact that the object moves at a constant speed around the circle. One full revolution around the circle corresponds to an angle of 2pi radians. So, the angle corresponding to time t would be:
theta = (2 * pi * t) / T
where T is the total time it takes to complete one revolution. In this case, the object takes T = 4 seconds to complete one revolution, as it travels from (5, 0) to (-5, 0) and back.
Using this equation, we can find the object's position at any time t.
Now let's find the object's speed:
The distance traveled in one revolution is the circumference of the circle, which is given by:
C = 2 * pi * r = 2 * pi * 5 = 10 * pi
Since the object takes T = 4 seconds to complete one revolution, we can find its speed:
Speed = Distance / Time = (10 * pi) / 4 = 5 * pi / 2 ≈ 7.85 meters per second
To find when the object returns to (5, 0), we can look for the values of t that correspond to x = 5 and y = 0.
When x = 5:
5 = 5 * cos((2 * pi * t) / 4)
cos((2 * pi * t) / 4) = 1
(2 * pi * t) / 4 = 0
t = 0
Therefore, the object returns to (5, 0) at t = 0.
To summarize, here are the object's positions for different values of t:
- When t = 0, the object is at (5, 0).
- When t = 1, the object is at approximately (3.54, 3.54).
- When t = 2, the object is at (0, 5).
- When t = 3, the object is at approximately (-3.54, 3.54).
- The object returns to (5, 0) at t = 0.
- The object's speed is approximately 7.85 meters per second.
To find the position of the object at different times (t = 2, t = 3, t = n) and the times when it returns to (5, 0), we will use the concepts of parametric equations for the motion along a circle.
1. Finding the equation of motion:
The equation x^2 + y^2 = 25 represents a circle with a radius of 5 units centered at the origin (0, 0). Since the object starts at (5, 0), we can rewrite the equation as:
(x - 5)^2 + y^2 = 0
When t = 0 seconds, the object is at (5, 0), and when t = 1 second, it is at (4, 3). We can represent the object's motion using the parametric equations for a circle:
x = 5 + r*cos(theta)
y = r*sin(theta)
Here, r is the radius of the circle (5), and theta is the angle at which the object is located on the circle. To find theta, we can use the following formula:
theta = arctan(y/x)
2. Finding the object's position at t = 2 and t = 3:
To find the position of the object when t = 2, we substitute t = 2 into the parametric equations:
x = 5 + r*cos(2*theta)
y = r*sin(2*theta)
Similarly, to find the position at t = 3, we substitute t = 3:
x = 5 + r*cos(3*theta)
y = r*sin(3*theta)
3. Finding the object's speed:
To find the object's speed, we need to calculate the distance traveled in one unit of time. Since the object moves at a constant speed, it will travel the same distance along the circle for each unit of time.
The circumference of a circle can be calculated using the formula:
C = 2*pi*r
The speed will be equal to the circumference divided by the time taken to complete one full revolution. In this case, it will be:
Speed = (2*pi*r) / T
4. Finding when the object returns to (5, 0):
To find when the object returns to (5, 0), we can set the value of theta equal to zero, as (5, 0) corresponds to theta = 0 on the circle. We solve the equation:
5 + r*cos(theta) = 5
r*cos(theta) = 0
cos(theta) = 0
The cosine function is zero at multiples of pi/2. Therefore, the object will return to (5, 0) when theta is equal to 0, pi/2, pi, 3pi/2, etc. We can use the formula for theta to find the corresponding values of t.
That's the explanation on how to approach the problem. Do you have any further questions?
in 1 second, it has moved through an angle of θ radians, where
tanθ = 3/4
so after n seconds, it will have moved through an angle of
Ø = n arctan(3/4)
Its speed is s = rθ m/s
it will be back at (5,0) when t = 2π/θ