A jet plane is flying at a constant altitude. At time t1 = 0, it has components of velocity vx = 88 m/s, vy = 115 m/s. At time t2 = 30s, the components are vx = -175 m/s, vy = 35 m/s.
a) Sketch the velocity vectors at t1 and t2. (figured that out on paper)
b) For this time interval calculate the components of the average acceleration, and c) the magnitude and direction of the average acceleration.
I think average acceleration (x, y) should be (-8.77, -2.67) because the x component = change in x components of velocity / change in time = (-175 - 88) / 30. And then the same respectively for y component of average acceleration.
This makes the magnitude the square root of (-8.77^2) + (-2.67^2) which = 9.17 m/s^2. Direction = inverse tangent of (-2.67 / -8.77) = 16.9, and then add 180 to put it in quadrant III.
So these answers make sense to me, but they contradict the answers in my textbook. I have had a problem with the answer key in my textbook before. I'm trying to check my work against the answer key before I submit it online, but I'm beginning to feel like the textbook answers can be misleading.
Am I correct?
b. V1 = 88 + 115i = 144.8m/s.[52.6o]. = Velocity at t1.
V2 = -175 + 35i = 178.5m/s[11.3o] N. of W. =178.5m/s[168.7o] CCW.
a = (V2 - V1) /(t2-t1) = (-175+35i -88-115i) /30 = -263 - 80i)/30 = -8.77- 2.67i m/s^2.
c. a = -8.77 - 2.67i = 9.17m/s^2[16.9o] S. of W. = 9.17m/s^2[196.9o] CCW.
My answers agree with yours.
Based on your calculations, it seems like you have made an error while calculating the x-component of the average acceleration. Let's go through the calculations again to determine the correct answers.
a) Since you have already sketched the velocity vectors at t1 and t2 on paper, we can move on to the next part.
b) To calculate the components of the average acceleration, we can use the formulas:
average acceleration in x-direction = (final velocity in x-direction - initial velocity in x-direction) / time interval
average acceleration in y-direction = (final velocity in y-direction - initial velocity in y-direction) / time interval
Using the given values:
Initial velocity in x-direction (t1) = 88 m/s
Final velocity in x-direction (t2) = -175 m/s
Initial velocity in y-direction (t1) = 115 m/s
Final velocity in y-direction (t2) = 35 m/s
Time interval = 30s
Average acceleration in x-direction = (-175 m/s - 88 m/s) / 30s = -8.43 m/s^2
Average acceleration in y-direction = (35 m/s - 115 m/s) / 30s = -2.67 m/s^2
c) To find the magnitude of the average acceleration, we can use the formula:
magnitude of average acceleration = √(average acceleration in x-direction)^2 + (average acceleration in y-direction)^2
Using the values calculated above:
magnitude of average acceleration = √((-8.43 m/s^2)^2 + (-2.67 m/s^2)^2) = √(70.9649 + 7.1289) = √78.0938 = 8.83 m/s^2
To find the direction of the average acceleration, we can use the formula:
direction = arctan(average acceleration in y-direction / average acceleration in x-direction)
Using the values calculated above:
direction = arctan(-2.67 m/s^2 / -8.43 m/s^2) ≈ 16.5°
Please note that inverse tangent values can vary based on calculator settings, so slight differences in the direction may occur.
Comparing these calculations with your previous answers, it is possible that the textbook answer key is incorrect. However, it is always a good idea to double-check or consult with your instructor for confirmation.
To calculate the components of average acceleration during the given time interval, we can use the formula:
Average acceleration (ax, ay) = (change in velocity in x direction / change in time, change in velocity in y direction / change in time).
Let's calculate the average acceleration:
For the x-component:
change in vx = (-175 m/s) - (88 m/s) = -263 m/s
change in time = 30 s
ax = (-263 m/s) / (30 s) = -8.77 m/s^2
For the y-component:
change in vy = (35 m/s) - (115 m/s) = -80 m/s
change in time = 30 s
ay = (-80 m/s) / (30 s) = -2.67 m/s^2
So your calculation for average acceleration (x, y) = (-8.77, -2.67) is correct.
Now, let's calculate the magnitude of the average acceleration using the formula:
Magnitude = sqrt(ax^2 + ay^2)
Magnitude = sqrt((-8.77 m/s^2)^2 + (-2.67 m/s^2)^2)
Magnitude ≈ 9.17 m/s^2
Your calculation for the magnitude of the average acceleration is also correct.
To find the direction of the average acceleration, we can use the inverse tangent (arctan) function. However, be careful with the signs. Based on the given components, the direction should be calculated as follows:
Direction = arctan(ay / ax)
Direction = arctan((-2.67 m/s^2) / (-8.77 m/s^2))
Direction ≈ 16.9 degrees
Since the y-component of acceleration is negative, the angle is in the fourth quadrant, which is equivalent to adding 180 degrees to the calculated angle.
Direction ≈ 16.9 degrees + 180 degrees ≈ 196.9 degrees
Therefore, your calculation for the direction of the average acceleration is correct.
If your answers contradict the answers in your textbook, it's possible that there might be an error in the textbook's answer key. It's always good to double-check your work and trust your own calculations if you believe they are correct.