Two particles are moving in straight lines. The displacement (meters) of particle 1 is given by the function s(t)= cos(4x), where t is in seconds. The displacement (meters) of particle 2 is given by the function s(t)= t^3/3-t^2/2 +2(t) , where t is in seconds. Find the first positive time at which the particles have the same velocity.
(Points : 1)
you want where s' is the same for both particles
-4sin(4x) = x^2-x+2
Better use a graphing utility of numeric method. See
http://www.wolframalpha.com/input/?i=-4sin%284x%29+%3D+x^2-x%2B2
Thanks :)
To find the first positive time at which the particles have the same velocity, we need to equate the derivatives of their displacement functions.
Let's start by finding the derivative of particle 1's displacement function, s(t) = cos(4t).
The derivative of cos(4t) with respect to t can be found using the chain rule of differentiation. Since the derivative of cos(x) is -sin(x), we have:
s'(t) = -sin(4t) * 4
Now, let's find the derivative of particle 2's displacement function, s(t) = t^3/3 - t^2/2 + 2t.
The derivative of t^3/3 with respect to t is (3t^2)/3 = t^2.
The derivative of -t^2/2 with respect to t is -2t/2 = -t.
The derivative of 2t with respect to t is 2.
Therefore, the derivative of s(t) = t^3/3 - t^2/2 + 2t is:
s'(t) = t^2 - t + 2
Now, we need to find the time at which the velocities of the two particles are equal. This happens when the derivatives of their displacement functions are equal. Therefore, we equate the derivatives:
-sin(4t) * 4 = t^2 - t + 2
To find the first positive time at which this equation holds true, we can solve it numerically or graphically. One common numerical method of solving equations is the Newton-Raphson method. For a graphical solution, we can plot both sides of the equation on a graphing calculator or software and find the intersection point.
Once the equation is solved, we can find the first positive time value at which the particles have the same velocity.