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

vx1 = 695 km/h = 193.06 m/s

vy1 = 440 km/h = 122.22 m/s
vx2 = 933 km/h = 259.17 m/s
vy2 = 465 km/h = 129.17 m/s

avarage ax = (vx2 - vx1)/7.12 s
= 9.29 m/s^2

average ay = (vy2 - vy1)/7.12s
= 0.98 m/s^2

I hope you can see where you made your error and proceed from there.

but it has to equal km/h^2 and i got those anwsers wrong

I am surprised they are asking for acceleration in unconventional units of km/h^2. You should recheck that.

If you want km/h^2 units, the time interval is
7.12 s = 1.978*10^-3 h

ax = (933-695)/1.9878*10^-3
= 3.02*10^4 km/h^2
so I still don't see how you got your answer of 104.8 in whatever units you are using

(933-695)/7.12- 4.85 = 238/2.27= 104.84 that's how i got that anwser...

You are quite right, I forgot to do the time subtraction for the interval. I also made a computational error in the division.

7.12 - 4.85 = 2.27 s = 6.306*10^-4 h

ax = 29.1 m/s^2
= 3.77*10^5 km/h^2

Here is what you did wrong:
To get acceleration in units of km/h^2, you divided the speed change in km/h by the time interval in SECONDS.

What you did gave you the average acceleration in (km per hour) per second

how did you get 29.1 to = 3.77* 10^5?

nevermind you converted it

b i got 3.96x10^4 km.h and i already did the magnitude which it 3.80x10^5

for the last question about the "for this time interval, find the direction of its average acceleration.
= ? countercloclock wise from + x axis

The direction of the avg. acceleration, clockwise from +x axis, is arctan ay/ax

You have the wrong dimensions following your ax. km.h is not a dimension of acceleration.

To solve this problem, we will use the average acceleration formula:

Average acceleration (ax) = (Vx2 - Vx1)/(t2 - t1)
Average acceleration (ay) = (Vy2 - Vy1)/(t2 - t1)

Given: Vx1 = 695 km/h, Vy1 = 440 km/h, t1 = 4.85 s
Vx2 = 933 km/h, Vy2 = 465 km/h, t2 = 7.12 s

(a) To find the x-component of the plane's average acceleration:
ax = (Vx2 - Vx1)/(t2 - t1)
= (933 km/h - 695 km/h)/(7.12 s - 4.85 s)
= 238 km/h / 2.27 s
= 104.81 km/h/s (rounded to two decimal places)

(b) To find the y-component of the plane's average acceleration:
ay = (Vy2 - Vy1)/(t2 - t1)
= (465 km/h - 440 km/h)/(7.12 s - 4.85 s)
= 25 km/h / 2.27 s
= 11 km/h/s (rounded to two decimal places)

Now, to find the magnitude of the average acceleration, you can use the Pythagorean theorem:
Magnitude of average acceleration = sqrt(ax^2 + ay^2)
= sqrt((104.81 km/h/s)^2 + (11 km/h/s)^2)
= sqrt(11033.6561 km^2/h^2/s^2 + 121 km^2/h^2/s^2)
= sqrt(11154.6561 km^2/h^2/s^2)
= 105.68 km/h/s (rounded to two decimal places)

To find the direction of average acceleration, you can use the inverse tangent function:
Direction = tan^(-1)(ay/ax)
= tan^(-1)(11 km/h/s / 104.81 km/h/s)
= tan^(-1)(0.105, rounded to three decimal places)
= 6.021 degrees

Therefore, the direction of the average acceleration is counterclockwise from the +x axis by an angle of approximately 6.021 degrees.