An asteroid heads for Earth at 12 km/s. In addition, a NASA space team is able to attach a rocket booster to the asteroid. The rocket booster fires for 10 min after which the asteroid is moving at 28 degrees to its original path to a speed of 20km/s. What is its average accleration for acceleartion at (x) and acceleartion at(y).
For acceleration (x) I got 12.9 m/s^2,but I am not sure about that or acceleartion at (y). Any help would be greatly appreciated.
acceleration= Vf-Vi /time
But draw the vector diagram, you see that Vf-Vi is the third vector in the vector triangle, and the law of cosines lets it be found...
(Vf-vi)^2=Vf^2+ Vi^2 - 2vf*Vi cos 28
acceleration then is (Vf-Vi) magnitude, divide by time in seconds.
I have no idea what you mean by this is my v initial 12km/s and my v final 20km/s
Nope, they have direction associated with them. Vf-Vi is a vector, it is NOT 20-12
To find the average acceleration for acceleration at (x) and acceleration at (y), we need to calculate the change in velocity in the x and y directions, and divide it by the total time taken.
Firstly, let's calculate the change in velocity in the x direction. We can use the initial and final velocities in the x direction.
Initial velocity in the x direction (u_x) = 12 km/s
Final velocity in the x direction (v_x) = 20 km/s
Change in velocity in the x direction (Δv_x) = v_x - u_x
= 20 km/s - 12 km/s
= 8 km/s
Since we are dealing with acceleration, we need to convert the change in velocity from km/s to m/s.
Δv_x = 8 km/s * (1000 m/1 km)
= 8000 m/s
Next, let's calculate the change in velocity in the y direction. We need to find the difference between the initial and final velocities in the y direction.
Since the asteroid is moving at an angle of 28 degrees to its original path, we can use trigonometry to find the change in velocity in the y direction.
Change in velocity in the y direction (Δv_y) = v_y - u_y
= v * sin(28°)
Given that v = 20 km/s, we can calculate Δv_y.
Δv_y = 20 km/s * sin(28°)
= 20 km/s * 0.4695
≈ 9.39 km/s
Again, we need to convert the change in velocity from km/s to m/s.
Δv_y = 9.39 km/s * (1000 m/1 km)
= 9390 m/s
Now, we have calculated the changes in velocity in both the x and y directions.
To find the average acceleration, we divide the changes in velocity by the time taken.
The time taken for the rocket booster to fire is 10 minutes, which is 600 seconds.
Average acceleration in the x direction (a_x) = Δv_x / t
= 8000 m/s / 600 s
≈ 13.33 m/s²
Average acceleration in the y direction (a_y) = Δv_y / t
= 9390 m/s / 600 s
≈ 15.65 m/s²
Therefore, the average acceleration for acceleration at (x) is approximately 13.33 m/s², and for acceleration at (y) it is approximately 15.65 m/s².