Two particles are moving in straight lines. The displacement (in meters) of particle 1 is given by the function e^(4cos(t)) , where t is in seconds. The displacement (in meters) of particle 2 is given by the function -(t^3)/(3) - (t^2)/(2) + 2 , where t is in seconds. Find the first positive time at which the particles have(approximately) the same velocity.
A.) t = 1.569 seconds
B.) t = 0 seconds
C.) t = 2.366 seconds
D.) t = 0.588 seconds
E.) t = 1.011 seconds
v1 = -4 sint e^4cost
v2 = -t^2 - t
so when does 4 sin t e^4 cos t = t^2+t ??
try approximately those answers to see when v1=v2
A 3.96 and 4.16
B 4 and 0 no
C the rest look pretty far off
try A accurately
4.0288 and 4.0307
It is A
Yes the answer is t=1.569
the first positive time which the two particles have the same velocity is at 1.569 seconds. checck this with nderv on your cal. easier to find the difference to find where its derivative is 0.
Well, well, well, it seems like we have a physics problem on our hands! Let's get cracking.
To find the first positive time at which the particles have the same velocity, we need to set their velocities equal to each other. Velocity is the derivative of displacement with respect to time, so let's differentiate the two given functions and set them equal to each other.
For particle 1, the velocity is given by the derivative of e^(4cos(t)), which is -4sin(t)e^(4cos(t)).
For particle 2, the velocity is given by the derivative of -(t^3)/(3) - (t^2)/(2) + 2, which is -t^2 - t.
Setting these two velocities equal to each other gives us the equation -4sin(t)e^(4cos(t)) = -t^2 - t.
Now, solving this equation analytically is a bit of a clown's task, so let's use some numerical methods or approximation techniques.
I've done some calculations, and it looks like the first positive time at which the particles have approximately the same velocity is around t = 1.569 seconds.
So, my dear friend, the answer is A.) t = 1.569 seconds.
Remember, physics problems love to play games with us, so it's always good to double-check the answer and make sure those pesky particles aren't fooling around.
To find the time at which the particles have approximately the same velocity, we need to find the time at which the derivatives of their displacement functions are approximately equal.
Let's find the derivatives of the displacement functions of both particles.
For particle 1, the displacement function is e^(4cos(t)). To find its derivative, we apply the chain rule:
d/dt [e^(4cos(t))] = e^(4cos(t)) * [-4sin(t)]
For particle 2, the displacement function is -(t^3)/(3) - (t^2)/(2) + 2. To find its derivative, we differentiate each term:
d/dt [-(t^3)/(3) - (t^2)/(2) + 2] = -3(t^2)/3 - 2(t)/2 + 0 = -t^2 - t
Now, we want to find the time when the derivatives are approximately equal. So, we set the two derivatives equal to each other and solve for t:
e^(4cos(t)) * [-4sin(t)] = -t^2 - t
To solve this equation, we can use numerical methods or de-arrange the terms and use trial and error. Let's rearrange the equation:
t^2 + t + [e^(4cos(t)) * 4sin(t)] = 0
Using numerical methods or trial and error, we find that the first positive solution for t is approximately 1.569 seconds.
Therefore, the correct answer is:
A.) t = 1.569 seconds
To find the first positive time at which the particles have approximately the same velocity, we need to equate their velocities and solve for t.
The velocity of particle 1 can be found by taking the derivative of its displacement function:
v1(t) = d/dt (e^(4cos(t))) = -4sin(t) * e^(4cos(t))
The velocity of particle 2 can be found by taking the derivative of its displacement function:
v2(t) = d/dt (-(t^3)/(3) - (t^2)/(2) + 2) = -t^2 - t
Now, we can equate the velocities of the particles:
-4sin(t) * e^(4cos(t)) = -t^2 - t
At this point, we can try to find the exact solution by solving this equation analytically, but it involves complex mathematical techniques which may not yield a simple expression for t. So, we will approximate the solution using numerical methods.
One way to approximate the solution is by graphing both velocity functions and finding the intersection point. You can use a graphing calculator or software to plot the two functions and find the point of intersection.
Another way is to use a numerical solver, such as the Newton-Raphson method, to find the approximate solution. The Newton-Raphson method involves making an initial guess for the value of t and then iteratively improving the guess until we reach a solution.
Here's how you can use an online Newton-Raphson calculator to find the approximate solution:
1. Go to an online numerical solver website that supports the Newton-Raphson method.
2. Enter the function -4*sin(t) * exp(4*cos(t)) - (-t^2 - t) = 0.
3. Specify a suitable initial guess for t. For example, you can start with t = 0.
4. Run the solver and it will provide an approximate solution.
After using a numerical solver, the approximate solution I found is t = 1.569 seconds.
Therefore, the answer is A.) t = 1.569 seconds.