A plane is flying in the horizontal plane. At time t1= 0 s with the components of its velocity are vx= 100 m/s and vy= 120 m/s. At time t2= 20 s the components are vx= 80 m/s and vy = 142 m/s.
a. What was the change in the direction of the velocity?
b. What is the change in the magnitude of the plane’s velocity during this time?
c. What are the components of the average acceleration for this interval?
d. What are the magnitude and direction of the average acceleration for this interval?
a. tanA1 = Vy/Vx = 120/100 = 1.20.
A1 = 50.2o = Direction.
Magnitude = Vx/cosA1=100/cos50.2 =156.2
m/s.
tanA2 = 142/80 = 1.775.
A2 = 60.6o = Direction.
Magnitude = 80/cos60.6 = 163 m/s.
60.6 - 50.2 = 10.4o = The change in
direction.
b. 163 - 156 = 7 m/s.
c. a(x) = (80-100)/20 = -1.0 m/s^2.
a(y) = (142-120)/20 = 1.1 m/s^2.
d, tanAr = 1.1/-1 = -1.10.
Ar = -47.7o = Reference angle.
A = 180 + (-47.7) = 132.3o=Direction.
Mag, = -1.0/cos132.3 = 1.49 m/s^2.
To find the answers to these questions, we need to apply basic principles of kinematics and vector analysis. Let's go through each question one by one:
a. What was the change in the direction of the velocity?
To find the change in direction of velocity, we need to calculate the angle between the initial and final velocity vectors. We can use trigonometry to do this. The formula to find the angle between two vectors is given by:
θ = arctan((y2 - y1) / (x2 - x1))
where (x1, y1) and (x2, y2) are the components of the initial and final velocity vectors, respectively.
In this case, using the given values:
θ = arctan((142 - 120) / (80 - 100))
θ = arctan(22 / -20)
θ ≈ -51.34 degrees (to two decimal places)
Therefore, the change in the direction of the velocity is approximately 51.34 degrees to the left (clockwise) from the positive x-axis.
b. What is the change in the magnitude of the plane's velocity during this time?
To find the change in magnitude of velocity, we calculate the difference between the initial and final magnitudes.
Initial magnitude = sqrt(vx^2 + vy^2) = sqrt((100)^2 + (120)^2) ≈ 155.56 m/s
Final magnitude = sqrt(vx^2 + vy^2) = sqrt((80)^2 + (142)^2) ≈ 164.92 m/s
Change in magnitude = Final magnitude - Initial magnitude = 164.92 m/s - 155.56 m/s ≈ 9.36 m/s
Therefore, the change in the magnitude of the plane's velocity during this time is approximately 9.36 m/s.
c. What are the components of the average acceleration for this interval?
The average acceleration can be calculated using the equation:
average acceleration = (change in velocity) / (time taken)
Change in velocity = (final velocity) - (initial velocity)
In this case, the initial and final velocities are given as components vx and vy. Therefore,
Change in velocity in the x-direction = vx2 - vx1 = 80 m/s - 100 m/s = -20 m/s
Change in velocity in the y-direction = vy2 - vy1 = 142 m/s - 120 m/s = 22 m/s
Therefore, the components of the average acceleration are -20 m/s in the x-direction and 22 m/s in the y-direction.
d. What are the magnitude and direction of the average acceleration for this interval?
The magnitude of the average acceleration can be calculated using Pythagoras' theorem:
Magnitude of average acceleration = sqrt((change in velocity in x-direction)^2 + (change in velocity in y-direction)^2)
= sqrt((-20)^2 + (22)^2)
= sqrt(400 + 484)
= sqrt(884)
≈ 29.73 m/s^2 (to two decimal places)
To find the direction, we use trigonometry. The direction can be determined using the angle:
θ = arctan((change in velocity in y-direction) / (change in velocity in x-direction))
= arctan(22 / -20)
≈ -48.37 degrees (to two decimal places)
The magnitude and direction of the average acceleration for this interval are approximately 29.73 m/s^2 and -48.37 degrees (to the left, clockwise) from the positive x-axis, respectively.