Some rental cars have a GPS unit installed, which allows the rental car company to check where you are at all times and thus also know your speed at any time. One of these rental cars is driven by an employee in the company's lot and, during the time interval from 0 to \(10 \mathrm{~s}\), is found to have a position vector as a function of time of $$ \begin{aligned} \vec{r}(t)=&\left((24.4 \mathrm{~m})-t(12.3 \mathrm{~m} / \mathrm{s})+t^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+t^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-t^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ a) What is the distance of this car from the origin of the coordinate system at \(t=5.00 \mathrm{~s} ?\) b) What is the velocity vector as a function of time? c) What is the speed at \(t=5.00 \mathrm{~s} ?\) Extra credit: Can you produce a plot of the trajectory of the car in the \(x y\) -plane?

Step by step solution

01

Evaluate position vector at \(t = 5.00 \mathrm{~s}\)

To find the position of the car at \(t = 5.00 \mathrm{~s}\), we need to substitute the value of t into the given position vector function \(\vec{r}(t)\): $$ \begin{aligned} \vec{r}(5.00 \mathrm{~s})=&\left((24.4 \mathrm{~m})-5(12.3 \mathrm{~m}\mathrm{s}^{-1})+5^{2}\left(2.43 \mathrm{~m}\mathrm{s}^{-2}\right)\right.\\ &\left.(74.4 \mathrm{~m})+5^{2}\left(1.80 \mathrm{~m}\mathrm{s}^{-2}\right)-5^{3}\left(0.130 \mathrm{~m}\mathrm{s}^{-3}\right)\right) \end{aligned} $$ Compute the result to obtain the position vector at \(t = 5.00 \mathrm{~s}\).

02

Calculate distance from the origin

Now that we have the position vector of the car at \(t = 5.00 \mathrm{~s}\), we can calculate the distance from the origin using the formula for the magnitude of a vector: $$ |\vec{r}| = \sqrt{x^{2} + y^{2}} $$ Substitute the components of the position vector and compute the result to find the distance from the origin at \(t = 5.00 \mathrm{~s}\).

03

Find the velocity vector function

In order to find the velocity vector as a function of time, we need to differentiate the position vector function \(\vec{r}(t)\) with respect to time t: $$ \vec{v}(t) = \frac{d\vec{r}}{dt} $$ Differentiate both the x and y components of the position vector function and express the result in vector form to obtain the velocity vector function \(\vec{v}(t)\).

04

Evaluate velocity at \(t = 5.00 \mathrm{~s}\)

Now we need to evaluate the velocity vector function at \(t = 5.00 \mathrm{~s}\) to find the velocity of the car at that time. Substitute the value of t into the velocity vector function \(\vec{v}(t)\) and compute the result to obtain the velocity at \(t = 5.00 \mathrm{~s}\).

05

Calculate speed at \(t = 5.00 \mathrm{~s}\)

Finally, we need to find the speed of the car at \(t = 5.00 \mathrm{~s}\). The speed can be calculated using the magnitude of the velocity vector: $$ v = |\vec{v}| $$ Substitute the components of the velocity vector found in Step 4 and compute the result to find the speed of the car at \(t = 5.00 \mathrm{~s}\).

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