Acceleration in 2 Dimensions-Vectors and Projectiles

[Snape1830]

1. An airplane is flying with a velocity of (150, 0) m/s. It is accelerated at 8.92 m/s2 in the x direction, and accelerated in negative y direction at 1.21 m/s2. What is the airplane's velocity at t = 3.6 s?2. My textbook says this is a relevant equation: average acceleration = change in velocity/change in time.
a_x=change in velocity_x/change in time
a_y=change in velocity_y/change in time.

Where v_x and v_y are the components of velocity.
3. So, I plugged the knows (time and acceleration) into the equations and got 32.1 m/s and 4.4 m/s. But I knew it was wrong because why would they give (150, 0) m/s?

How do I solve this problem?

 

[Doc Al]

You are finding the change in the velocity components. Use Δv = v_f - v_i.

 

[Snape1830]

Thanks Doc Al. The only issue is that I need two answers. (x,y) m/s.

 

[Doc Al]

The only issue is that I need two answers. (x,y) m/s.

And you have two components. What's the initial velocity x-component? y-component?

 

[asteriatic]

I would like to clarify that the given velocity (150, 0) m/s for the airplane is in vector form, with the first number representing the velocity in the x direction and the second number representing the velocity in the y direction. This means that the airplane is moving with a constant velocity of 150 m/s in the x direction and no velocity in the y direction.

To solve this problem, we can use the equations of motion for constant acceleration in two dimensions:

Vx = V0x + ax*t
Vy = V0y + ay*t

Where Vx and Vy are the final velocities in the x and y directions respectively, V0x and V0y are the initial velocities in the x and y directions respectively, ax and ay are the accelerations in the x and y directions respectively, and t is the time.

Substituting the given values, we get:

Vx = (150 m/s) + (8.92 m/s^2)(3.6 s) = 184.32 m/s
Vy = (0 m/s) + (-1.21 m/s^2)(3.6 s) = -4.356 m/s

Therefore, the airplane's velocity at t = 3.6 s is (184.32, -4.356) m/s.

As for the second question, the relevant equation for average acceleration is:

average acceleration = (change in velocity)/(change in time)

Therefore, the average acceleration in the x direction would be:

ax = (change in velocityx)/(change in time) = (184.32 m/s - 150 m/s)/(3.6 s) = 8.92 m/s^2

Similarly, the average acceleration in the y direction would be:

ay = (change in velocityy)/(change in time) = (-4.356 m/s - 0 m/s)/(3.6 s) = -1.21 m/s^2

I hope this clarifies the problem and helps you solve it correctly. As always, remember to pay attention to the units and use vector notation when dealing with quantities in two dimensions.