An aircraft flies at constant altitude (with respect to sea level) over the South Rim of the Grand Canyon (see figure below). Consider a coordinate system such that the positive x axis points to the east, and the positive y axis points north. The aircraft's initial position and velocity are 1330 m at an angle of 145° and 58.0 m/s at an angle of 55.0° where both angles are measured counterclockwise with respect to the positive x axis. The aircraft's acceleration is 3.8 m/s2 at an angle of 195° with respect to the positive x axis. (Express your answers in vector form.)

Question

An aircraft flies at constant altitude (with respect to sea level) over the South Rim of the Grand Canyon (see figure below). Consider a coordinate system such that the positive x axis points to the east, and the positive y axis points north. The aircraft's initial position and velocity are 1330 m at an angle of 145° and 58.0 m/s at an angle of 55.0° where both angles are measured counterclockwise with respect to the positive x axis. The aircraft's acceleration is 3.8 m/s2 at an angle of 195° with respect to the positive x axis. (Express your answers in vector form.)

(a) What is the velocity of the aircraft after 6.00 s have elapsed?
vf =
Incorrect:
m/s

(b) What is the position vector of the aircraft after 6.00 s have elapsed?

Answer 1

To find the velocity of the aircraft after 6.00 s, we need to use the formulas for horizontal and vertical components separately.

Given initial velocity (vi) and acceleration (a), we can use the formula:

vf = vi + at

For the horizontal component:

Given initial horizontal velocity (vi_x) = 58.0 m/s and horizontal acceleration (a_x) = 3.8 m/s^2

Using the formula vf_x = vi_x + a_x*t:

vf_x = 58.0 m/s + 3.8 m/s^2 * 6.00 s

vf_x = 58.0 m/s + 22.8 m/s

vf_x = 80.8 m/s

For the vertical component:

Given initial vertical velocity (vi_y) = 58.0 m/s and vertical acceleration (a_y) = 0 m/s^2 (since the aircraft flies at a constant altitude)

Using the formula vf_y = vi_y + a_y*t:

vf_y = 58.0 m/s + 0 m/s^2 * 6.00 s

vf_y = 58.0 m/s

Therefore, the velocity of the aircraft after 6.00 s is:

vf = (vf_x, vf_y) = (80.8 m/s, 58.0 m/s)

(a) vf = (80.8 m/s, 58.0 m/s)

To find the position vector of the aircraft after 6.00 s, we need to use the formula:

Δr = vi*t + 0.5*a*t^2

First, find the horizontal displacement (Δx):

Given initial horizontal position (xi) = 1330 m and initial horizontal velocity (vi_x) = 58.0 m/s

Using the formula Δx = xi + vi_x*t + 0.5*a_x*t^2:

Δx = 1330 m + 58.0 m/s * 6.00 s + 0.5 * 3.8 m/s^2 * (6.00 s)^2

Δx = 1330 m + 348 m + 68.4 m

Δx = 1746.4 m

For the vertical displacement (Δy):

Given initial vertical position (yi) = 1330 m and initial vertical velocity (vi_y) = 58.0 m/s

Using the formula Δy = yi + vi_y*t + 0.5*a_y*t^2:

Δy = 1330 m + 58.0 m/s * 6.00 s + 0.5 * 0 m/s^2 * (6.00 s)^2

Δy = 1330 m + 348 m + 0 m

Δy = 1678 m

Therefore, the position vector of the aircraft after 6.00 s is:

Δr = (Δx, Δy) = (1746.4 m, 1678 m)

(b) Δr = (1746.4 m, 1678 m)

Answer 2

To answer these questions, we need to break down the initial position, velocity, and acceleration vectors into their x and y components.

Given:
Initial position (xi, yi) = (1330 m, 0 m)
Initial velocity (vi, θv) = (58.0 m/s, 55.0°)
Acceleration (ai, θa) = (3.8 m/s^2, 195°)

(a) To find the velocity after 6.00 s, we need to find the final velocity vector (vf). We can use the equation:

vf = vi + a * t

First, let's find the x component of the initial velocity (vxi) and the x component of the acceleration (ax):

vxi = vi * cos(θv)
ax = a * cos(θa)

Substituting the known values:

vxi = 58.0 m/s * cos(55.0°)
ax = 3.8 m/s^2 * cos(195°)

Now, let's find the y component of the initial velocity (vyi) and the y component of the acceleration (ay):

vyi = vi * sin(θv)
ay = a * sin(θa)

Substituting the known values:

vyi = 58.0 m/s * sin(55.0°)
ay = 3.8 m/s^2 * sin(195°)

Now we can find the x component and y component of the final velocity (vxf, vyf):

vxf = vxi + ax * t
vyf = vyi + ay * t

Substituting the known values:

vxf = vxi + ax * 6.00 s
vyf = vyi + ay * 6.00 s

Finally, we can find the magnitude and direction of the final velocity using the Pythagorean theorem and inverse tangent function:

vf = sqrt(vxf^2 + vyf^2) m/s
θf = atan(vyf / vxf)

(b) To find the position vector after 6.00 s, we need to find the final position vector (rf). We can use the equation:

rf = ri + vi * t + 0.5 * a * t^2

First, let's find the x component of the initial position (xi) and the y component of the initial position (yi):

xi = 1330 m
yi = 0 m

Next, let's find the x component and y component of the final position (xf, yf):

xf = xi + (vxi * t) + (0.5 * ax * t^2)
yf = yi + (vyi * t) + (0.5 * ay * t^2)

Substituting the known values:

xf = xi + (vxi * 6.00 s) + (0.5 * ax * (6.00 s)^2)
yf = yi + (vyi * 6.00 s) + (0.5 * ay * (6.00 s)^2)

Therefore, the answers to the questions are:

(a) The final velocity of the aircraft after 6.00 s is vf = (vxf, vyf) m/s.
(b) The position vector of the aircraft after 6.00 s is rf = (xf, yf).