Same 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 10s, is found to have a position vector as a function of time of

Question

Same 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 10s, is found to have a position vector as a function of time of

r→(t) =( (24.4m)-t(12.3m/s)+t²(2.43m/s²),
(74.4m+t²(1.80m/s²)-t³(0.130m/s³) )
a) what is the distance of this car from the origin of the coordinate system at t=5.00s?
b) what is the velocity vector as a function of time?
c) what is the speed at t=5.00s?
Extra credit: Can you produce a plot of the trajectory of the car in the xy-plane?

Answer 1

a) do r(5)

b) dx/dt = u = -12.3 + 4.86 t
dy/dt = v = 3.6 t - .39 t^2

c) find sqrt (u^2+v^2) at t = 5

Answer 2

To find the distance of the car from the origin at t = 5.00s, we need to calculate the magnitude of the position vector at that time. The magnitude of a vector is given by the formula:

|v| = √(v₁² + v₂² + v₃²)

For the given position vector r(t), the magnitude can be calculated as follows:

|r(t)| = √((24.4m - t(12.3m/s) + t²(2.43m/s²))² + (74.4m + t²(1.80m/s²) - t³(0.130m/s³))²)

Substituting t = 5.00s into the position vector, we can calculate the distance from the origin:

|r(5.00s)| = √((24.4m - (5.00s)(12.3m/s) + (5.00s)²(2.43m/s²))² + (74.4m + (5.00s)²(1.80m/s²) - (5.00s)³(0.130m/s³))²)

Now, let's solve this equation step by step.

1. Evaluate (24.4m - (5.00s)(12.3m/s) + (5.00s)²(2.43m/s²)). This will give us the x-component of the position vector at t = 5.00s.
2. Evaluate (74.4m + (5.00s)²(1.80m/s²) - (5.00s)³(0.130m/s³)). This will give us the y-component of the position vector at t = 5.00s.
3. Square both the x-component and y-component obtained in steps 1 and 2, respectively.
4. Add the squared values obtained in step 3.
5. Take the square root of the sum obtained in step 4 to find the magnitude of the vector.
6. Round the answer to an appropriate number of significant figures to get the distance of the car from the origin at t = 5.00s.

To find the velocity vector as a function of time, we need to differentiate the position vector r(t) with respect to time. The velocity vector, v(t), is given by the derivative of the position vector:

v(t) = dr(t)/dt

For the given position function r(t), the velocity vector can be obtained as follows:

v(t) = (dr₁(t)/dt, dr₂(t)/dt)

Differentiating each component of r(t) with respect to t will give the velocity vector components.

To find the speed at t = 5.00s, we need to calculate the magnitude of the velocity vector at that time using the same formula:

|v(t)| = √(v₁² + v₂²)

Evaluate the magnitude of the velocity vector, substituting t = 5.00s into the velocity components obtained above.

To produce a plot of the trajectory of the car in the xy-plane, we can use graphing software or code libraries like Matplotlib in Python. We will plot the x and y components of the position vector against time to generate the trajectory.