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