Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
We are given that the position of an object if given by the following equations:
[tex]\begin{gathered} x=-3t^2+2t-4 \\ y=-2t^3+6t^2+1 \end{gathered}[/tex]To determine the velocity we will determine the derivative of each of the functions. For "x" we have:
[tex]x=-3t^2+2t-4[/tex]Finding the derivative with respect to time we get:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2+2t-4)[/tex]Now we distribute the derivative on the left side:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2)+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the first derivative we will use the rule:
[tex]\frac{d}{dt}(at^n)=ant^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the second derivative we use the rule:
[tex]\frac{d}{dt}(at)=a[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2-\frac{d}{dt}(4)[/tex]For the third derivative we use the rule:
[tex]\frac{d}{dt}(a)=0[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2[/tex]Now, since the velocity is the derivative with respect to time of the position and this is and we determine the derivative for the x-position what we have found is the velocity in the x-direction, therefore, we can write:
[tex]v_x=-6t^{}+2[/tex]Now we substitute the value of time, t = 1, we get:
[tex]\begin{gathered} v_x=-6(1)+2 \\ v_x=-6+2 \\ v_x=-4 \end{gathered}[/tex]Now we use derivate the function for "y":
[tex]\frac{dy}{dt}=\frac{d}{dt}(-2t^3+6t^2+1)[/tex]Using the same procedure as before we determine the derivative:
[tex]\frac{dy}{dt}=-6t^2+12t[/tex]This is the velocity in the y-direction:
[tex]v_y=-6t^2+12t[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} v_y=-6(1)^2+12(1) \\ v_y=6 \end{gathered}[/tex]Now, the speed is the magnitude of the velocity, the magnitude is given by:
[tex]v=\sqrt[]{v^2_x+v^2_y}[/tex]Substituting the values we get:
[tex]v=\sqrt[]{(-4)^2+6^2}[/tex]Solving the operations:
[tex]v=7.21[/tex]Therefore, the speed is 7.21 m/s.
To determine the acceleration we will determine the derivative of the formulas for velocities:
[tex]v_x=-6t^{}+2[/tex]Now we derivate with respect to time:
[tex]\frac{dv_x}{dt}=a_x=-6[/tex]Now we use the function for the velocity in the y-direction:
[tex]\frac{dv_y}{dt}=a_y=-12t^{}+12[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} a_y=-12(1)^{}+12 \\ a_y=0 \end{gathered}[/tex]Since the acceleration in the y-direction is zero, this means that the total acceleration is the acceleration in the x-direction, therefore, the magnitude of the acceleration is:
[tex]a=6[/tex]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
