Question

The position of an object is given by the formulasx = -3t² + 2t -4 and y = -2t³ + 6t² +1A) What is the speed at t = 1sB) What is the acceleration at t = 1s

Answer 1

We are given that the position of an object if given by the following equations:

$\begin{gathered} x=-3t^2+2t-4 \\ y=-2t^3+6t^2+1 \end{gathered}$

To determine the velocity we will determine the derivative of each of the functions. For "x" we have:

$x=-3t^2+2t-4$

Finding the derivative with respect to time we get:

$\frac{dx}{dt}=\frac{d}{dt}(-3t^2+2t-4)$

Now we distribute the derivative on the left side:

$\frac{dx}{dt}=\frac{d}{dt}(-3t^2)+\frac{d}{dt}(2t)-\frac{d}{dt}(4)$

For the first derivative we will use the rule:

$\frac{d}{dt}(at^n)=ant^{n-1}$

Applying the rule we get:

$\frac{dx}{dt}=-6t^{}+\frac{d}{dt}(2t)-\frac{d}{dt}(4)$

For the second derivative we use the rule:

$\frac{d}{dt}(at)=a$

Applying the rule we get:

$\frac{dx}{dt}=-6t^{}+2-\frac{d}{dt}(4)$

For the third derivative we use the rule:

$\frac{d}{dt}(a)=0$

Applying the rule we get:

$\frac{dx}{dt}=-6t^{}+2$

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:

$v_x=-6t^{}+2$

Now we substitute the value of time, t = 1, we get:

$\begin{gathered} v_x=-6(1)+2 \\ v_x=-6+2 \\ v_x=-4 \end{gathered}$

Now we use derivate the function for "y":

$\frac{dy}{dt}=\frac{d}{dt}(-2t^3+6t^2+1)$

Using the same procedure as before we determine the derivative:

$\frac{dy}{dt}=-6t^2+12t$

This is the velocity in the y-direction:

$v_y=-6t^2+12t$

Now we substitute the value of t = 1:

$\begin{gathered} v_y=-6(1)^2+12(1) \\ v_y=6 \end{gathered}$

Now, the speed is the magnitude of the velocity, the magnitude is given by:

$v=\sqrt[]{v^2_x+v^2_y}$

Substituting the values we get:

$v=\sqrt[]{(-4)^2+6^2}$

Solving the operations:

$v=7.21$

Therefore, the speed is 7.21 m/s.

To determine the acceleration we will determine the derivative of the formulas for velocities:

$v_x=-6t^{}+2$

Now we derivate with respect to time:

$\frac{dv_x}{dt}=a_x=-6$

Now we use the function for the velocity in the y-direction:

$\frac{dv_y}{dt}=a_y=-12t^{}+12$

Now we substitute the value of t = 1:

$\begin{gathered} a_y=-12(1)^{}+12 \\ a_y=0 \end{gathered}$

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:

$a=6$