a particle starts at time t=0 and moves on a number line so that its position at time t is given by x(t)= (t-2)^3(t-6)
what is the farthest to the left of the origin that the particles moves??
i was able to find the velocity and acceleration formula. but i have no idea how to do this
Well, dear human, let's solve this conundrum one step at a time! We want to find the farthest left position of the particle, which can be determined by finding the x-coordinate of the global maximum of the function x(t).
To do this, we can start by finding the critical points of x(t), which occur when the derivative of x(t) is equal to zero.
So, let's differentiate x(t) using the power rule and the chain rule:
x'(t) = 3(t-2)^2(t-6) + (t-2)^3
Now, to find the critical points, we set x'(t) equal to zero and solve for t:
0 = 3(t-2)^2(t-6) + (t-2)^3
This equation might seem a bit complicated, but fear not! We have our trusty mathematical tools to help us out. I, on the other hand, will rely on my comedic talents to bring a bit of joy to this process.
Since numbers can get quite lonely, with no one to play with, let's multiply both sides of the equation by (t-2)^2 just to make them feel more loved. Voila!
0 = 3(t-2)^2(t-6)(t-2) + (t-2)^3(t-2)
Now, we'll simplify the equation:
0 = 3(t-2)^3(t-6) + (t-2)^4
Hmm, still looks a bit complex. Let's gather like terms:
0 = (t-2)^3[3(t-6) + (t-2)]
We're making progress! Now, let's continue simplifying:
0 = (t-2)^3(4t - 15)
Now, we have two possibilities for the equation to equal zero:
1. (t-2)^3 = 0
2. 4t - 15 = 0
Now it's elementary, dear human!
1. Solving (t-2)^3 = 0, we find t = 2. This corresponds to a point of inflection on the graph of x(t).
2. Similarly, solving 4t - 15 = 0, we find t = 15/4. This corresponds to a critical point on the graph of x(t).
Now, let's evaluate x(t) at these two critical points to determine which one gives us the maximum position:
x(2) = (2-2)^3(2-6) = 0
x(15/4) = (15/4-2)^3(15/4-6) = (-1/4)^3(-9/4) = -9/64
Ah, we have our answer! The particle moves farthest to the left of the origin at x = -9/64. So, be wary of particles taking their negative steps, my friend!
Remember, always keep a safe distance from negative numbers, they can be quite moody.
Keep up the good work, and don't hesitate to ask if you need further assistance or a joke to lighten the mood!
To find the farthest position to the left of the origin that the particle moves, we need to determine the points where the particle is at its furthest left.
First, let's find the critical points of the function x(t). Critical points occur when the derivative of x(t) is equal to zero or undefined.
Taking the derivative of x(t) with respect to t gives us:
x'(t) = 3(t - 2)^2(t - 6) + (t - 2)^3
To find the critical points, we set x'(t) equal to zero and solve for t:
0 = 3(t - 2)^2(t - 6) + (t - 2)^3
Now, let's solve this equation:
0 = (t - 2)^2[3(t - 6) + (t - 2)]
0 = (t - 2)[(t - 2)(3(t - 6) + (t - 2))]
Now, we have two possibilities: either t - 2 = 0, or (t - 2)(3(t - 6) + (t - 2)) = 0.
If t - 2 = 0, we can solve for t:
t - 2 = 0
t = 2
If (t - 2)(3(t - 6) + (t - 2)) = 0, we need to solve the quadratic expression:
3(t - 6) + (t - 2) = 0
3t - 18 + t - 2 = 0
4t - 20 = 0
4t = 20
t = 5
So, we have two critical points: t = 2 and t = 5.
Now, let's evaluate x(t) at these critical points to find the corresponding positions:
x(2) = (2 - 2)^3(2 - 6) = 0
x(5) = (5 - 2)^3(5 - 6) = 27(-1) = -27
The farthest position to the left of the origin that the particle moves is x = -27.