At an air show, a jet plane has velocity components vx = 695 km/h and vy = 440 km/h at time 4.85s and vx= 933km/h and vy = 465 km/h at time 7.12 s. For this time interval, find (a) the x and y components of the plane’s average acceleration and (b) the magnitude and direction of its average acceleration
find (a) the x and y components of the plane’s average acceleration and (b) the magnitude and direction of its average acceleration
d) for this time interval, find the direction of its average acceleration?
=? counterclockwise from + x axis
im having a hard time solving this problem
There is nothing hard here.
ax= (Vxf-Vxi)/timedifference
(933-695)/(7.12-4.85) solve that, then do the same for
ay.
b) magnitude: sqrt(ax^2+ay^2)
direction: arctan ay/ax
ax i got 104.8 but its wrong and i rounded it to 105
To solve this problem, we can use the average acceleration formula:
Average acceleration (a) = (Final velocity - Initial velocity) / (Final time - Initial time)
Given:
Initial velocity (vx1) = 695 km/h
Initial velocity (vy1) = 440 km/h
Final velocity (vx2) = 933 km/h
Final velocity (vy2) = 465 km/h
Initial time (t1) = 4.85 s
Final time (t2) = 7.12 s
Step 1: Calculate the x-component of the average acceleration (ax):
ax = (vx2 - vx1) / (t2 - t1)
= (933 km/h - 695 km/h) / (7.12 s - 4.85 s)
= 238 km/h / 2.27 s
= 105.079 km/h/s
Step 2: Calculate the y-component of the average acceleration (ay):
ay = (vy2 - vy1) / (t2 - t1)
= (465 km/h - 440 km/h) / (7.12 s - 4.85 s)
= 25 km/h / 2.27 s
= 11.013 km/h/s
Therefore, the x-component of the average acceleration is 105.079 km/h/s and the y-component of the average acceleration is 11.013 km/h/s.
To find the magnitude and direction of the average acceleration, we can use the Pythagorean theorem and trigonometry.
Step 3: Calculate the magnitude of the average acceleration (|a|):
|a| = sqrt(ax^2 + ay^2)
= sqrt((105.079 km/h/s)^2 + (11.013 km/h/s)^2)
= sqrt(11071.92 km^2/h^2/s^2 + 121.34 km^2/h^2/s^2)
≈ 105.26 km/h/s
Step 4: Calculate the direction of the average acceleration (θ):
θ = arctan(ay / ax)
= arctan(11.013 km/h/s / 105.079 km/h/s)
= arctan(0.105)
≈ 5.99 degrees
Therefore, the magnitude of the average acceleration is approximately 105.26 km/h/s, and the direction is approximately 5.99 degrees counterclockwise from the +x axis.
To solve this problem, we need to find the average acceleration of the jet plane in both x and y directions and then calculate its magnitude and direction.
a) To find the average acceleration in the x-direction, we need to use the formula:
average acceleration (ax) = (final velocity (vxf) - initial velocity (vxi)) / (final time (tf) - initial time (ti))
Plugging in the given values:
vxf = 933 km/h
vxi = 695 km/h
tf = 7.12 s
ti = 4.85 s
ax = (933 km/h - 695 km/h) / (7.12 s - 4.85 s)
= 238 km/h / 2.27 s
Calculating the average acceleration in the x-direction gives us:
ax = 104.84 km/h/s (rounded to two decimal places)
Similarly, we can find the average acceleration in the y-direction using the same formula:
average acceleration (ay) = (final velocity (vyf) - initial velocity (vyi)) / (final time (tf) - initial time (ti))
Plugging in the given values:
vyf = 465 km/h
vyi = 440 km/h
tf = 7.12 s
ti = 4.85 s
ay = (465 km/h - 440 km/h) / (7.12 s - 4.85 s)
= 25 km/h / 2.27 s
Calculating the average acceleration in the y-direction gives us:
ay = 11.01 km/h/s (rounded to two decimal places)
b) To find the magnitude of the average acceleration, we can use the Pythagorean theorem:
magnitude (a) = √(ax^2 + ay^2)
Plugging in the calculated values:
ax = 104.84 km/h/s
ay = 11.01 km/h/s
Calculating the magnitude of the average acceleration gives us:
magnitude (a) = √(104.84^2 + 11.01^2)
= √(10991.1056 + 121.0201)
= √11112.1257
≈ 105.49 km/h/s (rounded to two decimal places)
c) To find the direction of the average acceleration, we can use the inverse tangent function:
direction (θ) = tan^(-1)(ay / ax)
Plugging in the calculated values:
ay = 11.01 km/h/s
ax = 104.84 km/h/s
Calculating the direction of the average acceleration gives us:
direction (θ) = tan^(-1)(11.01 / 104.84)
= tan^(-1)(0.1051)
≈ 6.010° (rounded to two decimal places)
Therefore, the x and y components of the plane's average acceleration are 104.84 km/h/s and 11.01 km/h/s, respectively. The magnitude of the average acceleration is approximately 105.49 km/h/s, and its direction is approximately 6.01° counterclockwise from the +x axis.