A meteor streaking through the night sky is located with radar. At point A its coordinates are (5.20km,1.20km), and 1.16s later its has moved to point B with coordinates (6.08km, 0.900km ).
a) find the x comonent of it average velocity between A and B. .759km/s
b) y= .259 km/s
c) magintude= .801
****the only problem im having trouble on is this....
Find the direction of its average velocity between these two points.
=? counterwise from axis
i still don't understand how to solve it ^
and solving this part D for this problem..
At an air show, a jet plane has velocity components Vx=695km and Vx =440km at time 4.85s and Vy=933 and Vy=465km at time 7.12s .
A) Vx= 3.77x10^5
B) Vy= 3.96 x10^4
C) magnitude= 3.80x10^5
D) for this time interval, find the direction of its average acceleration.
= ? counterclock wise from x axis
average velocity= change of displacement/time
you know the final and original positions..
changedisplacement= sqrt ((x2-x1)^2 + (y2-y1)^2)
second question. similar solution, only use intial and final velocityies to get average acceleration
those are the right anwsers im just having a hard time doing part D about the counterclock wise for both problem i tried doing tan^-1(vy/vx) and its giving me a wrong anwser
To find the direction of the average velocity between points A and B, we can use trigonometry. The direction can be represented as an angle with respect to the positive x-axis.
1. Determine the change in x-coordinate and change in y-coordinate between points A and B:
Δx = 6.08 km - 5.20 km = 0.88 km
Δy = 0.900 km - 1.20 km = -0.3 km
2. Calculate the angle using the inverse tangent function (tan⁻¹):
θ = tan⁻¹(Δy/Δx)
θ = tan⁻¹((-0.3 km)/(0.88 km))
θ ≈ -19.90°
The direction of the average velocity is approximately -19.90° counterclockwise from the x-axis.
Regarding solving part D for the second problem:
To find the direction of average acceleration, we need to consider the change in velocity components ΔVx and ΔVy over the given time interval.
1. Determine the change in Vx and Vy:
ΔVx = Vxf - Vxi = 440 km - 695 km = -255 km
ΔVy = Vyf - Vyi = 465 km - 933 km = -468 km
2. Calculate the angle using the inverse tangent function (tan⁻¹):
θ = tan⁻¹(ΔVy/ΔVx)
θ = tan⁻¹((-468 km)/(-255 km))
θ ≈ 60.25°
The direction of the average acceleration is approximately 60.25° counterclockwise from the x-axis.
To find the direction of the average velocity between two points, you can use trigonometry. First, calculate the change in x and y components of the velocity.
For this problem:
Δx = xB - xA = 6.08km - 5.20km = 0.88km
Δy = yB - yA = 0.900km - 1.20km = -0.3km
The negative sign in Δy indicates that the meteor is moving downwards in the y-direction.
Next, find the angle θ that the velocity vector makes with the positive x-axis using the inverse tangent function:
θ = arctan(Δy/Δx) = arctan((-0.3km)/(0.88km)) ≈ -19.4 degrees
The angle -19.4 degrees indicates that the direction of the average velocity is 19.4 degrees counter-clockwise from the positive x-axis.
For part D of the second problem, to find the direction of its average acceleration, you can follow the same approach.
First, calculate the change in x and y components of the velocity:
ΔVx = Vx(final) - Vx(initial) = 440km - 695km = -255km
ΔVy = Vy(final) - Vy(initial) = 465km - 933km = -468km
Again, the negative signs indicate that the jet plane is decelerating in both x and y directions.
Next, find the angle θ that the acceleration vector makes with the positive x-axis using the inverse tangent function:
θ = arctan(ΔVy/ΔVx) = arctan((-468km)/(-255km)) ≈ arctan(1.835) ≈ 61.19 degrees
The angle 61.19 degrees indicates that the direction of the average acceleration is 61.19 degrees counter-clockwise from the positive x-axis.