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
and for question (a) fo x i got 104.8 i got that wrong plus i even rounded it to 150 still wroung
Xf = 933000 m/h / 3600 s = 259.2 m/s.
Xo = 695000 m/h / 3600s = 193.1 m/s.
Yf = 465000 m/h / 3600 s = 129.2 m/s.
Yo = 440000 m/h / 3600 s = 122.2 m/s.
HOR.: (a) a == Xf-Xo) / (Tf-To),
a=(259.2-193.1) / (7.12-4.85)=29.1m/s^2
VER.: a = (Yf-Yo) / (Tf-To),
a = (129.2-122.2)/(7.12-4.85)=3.08m/s^2
b. tanA = Y/X = 3.08 / 29.1 = 0.10597.
A = 6.05 Deg.
a(avg)=X/cosA=29.1 / cos6.05=29.3m/s^2
@ 6.05 Deg.,CCW.
NOTE: You forgot to change unit of velocity to m/s.
To solve this problem, you can use the equations of motion for acceleration and velocity. Let's break it down step by step:
(a) The x and y components of the average acceleration can be calculated using the following formulas:
Average acceleration in the x direction = (final velocity in x - initial velocity in x) / (time interval)
Average acceleration in the y direction = (final velocity in y - initial velocity in y) / (time interval)
Given:
Initial velocity in x (vx) = 695 km/h
Final velocity in x (vx) = 933 km/h
Initial velocity in y (vy) = 440 km/h
Final velocity in y (vy) = 465 km/h
Time interval (Δt) = 7.12s - 4.85s = 2.27s
Now plug these values into the formulas:
Average acceleration in the x direction = (933 km/h - 695 km/h) / 2.27s
= 98.24 km/h/s (rounding to two decimal places)
Average acceleration in the y direction = (465 km/h - 440 km/h) / 2.27s
= 10.97 km/h/s (rounding to two decimal places)
Therefore, the x and y components of the plane's average acceleration are approximately 98.24 km/h/s and 10.97 km/h/s, respectively.
(b) The magnitude of the average acceleration can be found using the Pythagorean theorem:
Magnitude of average acceleration = sqrt((average acceleration in x)^2 + (average acceleration in y)^2)
Plugging in the values:
Magnitude of average acceleration = sqrt((98.24 km/h/s)^2 + (10.97 km/h/s)^2)
= sqrt(9669.44 km^2/h^2/s^2 + 120.44 km^2/h^2/s^2)
= sqrt(9789.88 km^2/h^2/s^2)
= 99 km/h/s (approximately, rounding to the nearest whole number)
The magnitude of the average acceleration is 99 km/h/s.
(c) The direction of the average acceleration can be determined using trigonometry. To find the direction, use the following equation:
Direction of average acceleration = atan(average acceleration in y / average acceleration in x)
Plugging in the values:
Direction of average acceleration = atan(10.97 km/h/s / 98.24 km/h/s)
= atan(0.1116)
Using a calculator, we find that the arctan of 0.1116 is approximately 6.39 degrees (rounding to two decimal places).
Therefore, for this time interval, the direction of the average acceleration is approximately 6.39 degrees counterclockwise from the +x axis.