A jet plane is flying at a constant altitude. At time t1=0 it has components of velocity Vx=90m/s, Vy=110m/s.At time t2=30.0s the components are Vx=-170m/s, Vy=40m/s.a).Sketch velocity vectors at t1 and t2.How do these two vectors differ?. For this time interval calculate b). The components of the average acceleration, and b). The magnitude and direction of the average acceleration.

Just do them separate first ax = 90-170/.5, etc

Then use pythag and tan-1 for the direction.

a) To sketch the velocity vectors at t1 and t2, we need to plot the vectors with their corresponding magnitudes and directions.

At time t1 = 0, the components of velocity are Vx = 90 m/s (positive x-axis direction) and Vy = 110 m/s (positive y-axis direction). So, the velocity vector at t1 can be represented as an arrow pointing in the direction of (90, 110).

At time t2 = 30.0s, the components of velocity are Vx = -170 m/s (negative x-axis direction) and Vy = 40 m/s (positive y-axis direction). So, the velocity vector at t2 can be represented as an arrow pointing in the direction of (-170, 40).

The two vectors differ in both magnitude and direction. The velocity vector at t1 has greater magnitudes in both x and y directions compared to the velocity vector at t2. Additionally, the direction of the velocity vector at t1 is towards the 1st quadrant, while the direction of the velocity vector at t2 is towards the 2nd quadrant.

b) To calculate the components of average acceleration, we can use the formula:

average acceleration (ax, ay) = (change in velocity in x direction, change in velocity in y direction) / (change in time)

Given:
t1 = 0
t2 = 30.0s
Vx1 = 90 m/s
Vy1 = 110 m/s
Vx2 = -170 m/s
Vy2 = 40 m/s

The change in velocity in the x direction is Vx2 - Vx1 = -170 m/s - 90 m/s = -260 m/s.
The change in velocity in the y direction is Vy2 - Vy1 = 40 m/s - 110 m/s = -70 m/s.
The change in time is t2 - t1 = 30.0s - 0s = 30.0s.

Therefore, the components of average acceleration are:
ax = (-260 m/s) / (30.0s) = -8.67 m/s^2 (in the x direction)
ay = (-70 m/s) / (30.0s) = -2.33 m/s^2 (in the y direction)

c) To calculate the magnitude and direction of average acceleration, we can use the Pythagorean theorem and trigonometry.

The magnitude of average acceleration (a) can be calculated as:
a = sqrt(ax^2 + ay^2) = sqrt((-8.67 m/s^2)^2 + (-2.33 m/s^2)^2).
a ≈ 9.02 m/s^2

The direction of average acceleration (θ) can be calculated as:
θ = atan(ay / ax) = atan((-2.33 m/s^2) / (-8.67 m/s^2)).
θ ≈ 15.7 degrees (approximated to 1 decimal place)

Therefore, the magnitude of the average acceleration is approximately 9.02 m/s^2, and the direction is approximately 15.7 degrees.

To sketch the velocity vectors at t1 and t2, we can use a coordinate system with the x-axis representing the horizontal direction (Vx) and the y-axis representing the vertical direction (Vy).

At t1 = 0, the jet plane has a velocity vector V1 = (Vx1, Vy1) = (90 m/s, 110 m/s). Draw an arrow starting from the origin and extending 90 units to the right (Vx) and 110 units upwards (Vy).

At t2 = 30.0 s, the jet plane has a velocity vector V2 = (Vx2, Vy2) = (-170 m/s, 40 m/s). Draw an arrow starting from the origin and extending 170 units to the left (Vx) and 40 units upwards (Vy).

To calculate the components of the average acceleration for this time interval, we need to use the formula:

Average acceleration = (Change in velocity) / (Time interval)

Change in velocity (ΔV) = V2 - V1

ΔVx = Vx2 - Vx1 = -170 m/s - 90 m/s = -260 m/s
ΔVy = Vy2 - Vy1 = 40 m/s - 110 m/s = -70 m/s

Time interval (Δt) = t2 - t1 = 30.0 s - 0 s = 30.0 s

Average acceleration = (ΔVx / Δt, ΔVy / Δt)
= (-260 m/s / 30.0 s, -70 m/s / 30.0 s)
= (-8.67 m/s², -2.33 m/s²)

So the components of the average acceleration are approximately (-8.67 m/s², -2.33 m/s²).

To calculate the magnitude and direction of the average acceleration, we can use the Pythagorean theorem and trigonometric functions:

Magnitude of the average acceleration (|a|) = sqrt((ΔVx)^2 + (ΔVy)^2)
= sqrt((-260 m/s)^2 + (-70 m/s)^2)
= sqrt(67600 m²/s² + 4900 m²/s²)
= sqrt(72500 m²/s²)
= 269.26 m/s² (rounded to two decimal places)

Direction of the average acceleration = arctan(ΔVy / ΔVx)
= arctan((-70 m/s) / (-260 m/s))
= arctan(0.27)
≈ 0.26 radians (rounded to two decimal places)

So the magnitude of the average acceleration is approximately 269.26 m/s² and the direction is approximately 0.26 radians (or about 14.90 degrees) from the negative x-axis.