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

rarrow(t) =
(24.4 m) − t(12.3 m/s) + t2(2.43 m/s2), (74.4 m) + t2(1.80 m/s2) − t3(0.130 m/s3)
.
(a) What is the distance of this car from the origin of the coordinate system at t = 5.89 s?
1

(b) What is the velocity vector as a function of time?
varrowitalic(t) = m/s

(c) What is the speed at t = 5.89 s?

To answer these questions, we need to calculate the distance, velocity vector, and speed at t = 5.89 s based on the given position vector.

(a) Distance from the origin at t = 5.89 s:
To find the distance from the origin, we need to calculate the magnitude of the position vector at t = 5.89 s.
The position vector is given by:

r⃗(t) = (24.4 m − t(12.3 m/s) + t^2(2.43 m/s^2), 74.4 m + t^2(1.80 m/s^2) − t^3(0.130 m/s^3))

Substitute t = 5.89 s into the position vector:

r⃗(5.89) = (24.4 m − 5.89(12.3 m/s) + (5.89)^2(2.43 m/s^2), 74.4 m + (5.89)^2(1.80 m/s^2) − (5.89)^3(0.130 m/s^3))

Evaluate the expression to get the position vector at t = 5.89 s and then calculate its magnitude to find the distance from the origin.

(b) Velocity vector as a function of time:
To find the velocity vector, we need to take the derivative of the position vector with respect to time.

v⃗(t) = dr⃗/dt

Differentiate each component of the position vector and simplify to get the velocity vector.

(c) Speed at t = 5.89 s:
To find the speed, we need to calculate the magnitude of the velocity vector at t = 5.89 s.

|v⃗(5.89)| = magnitude of the velocity vector at t = 5.89 s

Substitute t = 5.89 s into the velocity vector and calculate its magnitude to find the speed.

Note: Make sure to perform the necessary calculations using the given values and units to get accurate answers.

To solve this problem, we'll break it down step by step.

(a) To find the distance of the car from the origin at t = 5.89 s, we need to find the magnitude of the position vector at that time.

Given the position vector as a function of time: r(t) = (24.4 m) - t(12.3 m/s) + t^2(2.43 m/s^2), (74.4 m) + t^2(1.80 m/s^2) - t^3(0.130 m/s^3)

Substituting t = 5.89 s into the position vector, we have:

r(5.89) = (24.4 m) - (5.89 s)(12.3 m/s) + (5.89 s)^2(2.43 m/s^2), (74.4 m) + (5.89 s)^2(1.80 m/s^2) - (5.89 s)^3(0.130 m/s^3)

Evaluating these expressions, we find:

r(5.89) = (-44.737 m, 100.647 m)

The distance from the origin is the magnitude of this position vector:

| r(5.89) | = √((-44.737 m)^2 + (100.647 m)^2) = 109.239 m

So, the distance of the car from the origin at t = 5.89 s is approximately 109.239 m.

(b) The velocity vector as a function of time is obtained by taking the derivative of the position vector with respect to time:

v(t) = dr(t)/dt

Given the position vector as a function of time: r(t) = (24.4 m) - t(12.3 m/s) + t^2(2.43 m/s^2), (74.4 m) + t^2(1.80 m/s^2) - t^3(0.130 m/s^3)

Differentiating each component of the position vector with respect to time, we find:

v(t) = (-12.3 m/s + 2t(2.43 m/s^2), 2t(1.80 m/s^2) - 3t^2(0.130 m/s^3))

Simplifying this expression:

v(t) = (-12.3 m/s + 4.86t m/s, 3.6t m/s - 0.39t^2 m/s)

So, the velocity vector as a function of time is v(t) = (-12.3 m/s + 4.86t m/s, 3.6t m/s - 0.39t^2 m/s).

(c) The speed at t = 5.89 s is the magnitude of the velocity vector at that time.

Substituting t = 5.89 s into the velocity vector, we have:

v(5.89) = (-12.3 m/s + 4.86(5.89) m/s, 3.6(5.89) m/s - 0.39(5.89)^2 m/s)

Evaluating these expressions, we find:

v(5.89) = (18.032 m/s, 10.383 m/s)

The speed is the magnitude of this velocity vector:

|v(5.89)| = √((18.032 m/s)^2 + (10.383 m/s)^2) = 20.952 m/s

So, the speed at t = 5.89 s is approximately 20.952 m/s.