Solved: Kinematics Problem: Final Velocity

[aaronfue]

Homework Statement

A hovercraft is traveling 3m/s at 180 deg. The fan of the hovercraft causes a=4m/s² at 50° for 10 seconds.

Homework Equations

What is the final velocity?

The Attempt at a Solution

v = 3cos180 - 4cos50*10?

 

[gneill]

Homework Statement

A hovercraft is traveling 3m/s at 180 deg. The fan of the hovercraft causes a=4m/s² at 50° for 10 seconds.

Homework Equations

What is the final velocity?

The Attempt at a Solution

v = 3cos180 - 4cos50*10?

The velocity and acceleration should be vectors with both x and y components. Your attempt doesn't appear to deal with the y-component of the acceleration and its contribution to the final velocity (which will be a vector).

 

[aaronfue]

The velocity and acceleration should be vectors with both x and y components. Your attempt doesn't appear to deal with the y-component of the acceleration and its contribution to the final velocity (which will be a vector).

So do I have to use the same equation with 4sin50° to get the velocity in the y direction, then I can take the magnitude of both v_x & v_y to get the final velocity?

Do I have to do the same for the velocity using 3sin180?

 

[gneill]

So do I have to use the same equation with 4sin50° to get the velocity in the y direction, then I can take the magnitude of both v_x & v_y to get the final velocity?

Do I have to do the same for the velocity using 3sin180?

Sure, sounds good.

A note on terminology: In general, "velocity" is a vector which has both speed and direction. The magnitude of a velocity vector is the speed. Speed is a scalar quantity with no particular associated direction.

 

[aaronfue]

Sure, sounds good.

A note on terminology: In general, "velocity" is a vector which has both speed and direction. The magnitude of a velocity vector is the speed. Speed is a scalar quantity with no particular associated direction.

Ok. I calculated the velocity like so:

v_x = 3cos180 + 4cos50*10 = 22.71 m/s
v_y = 3sin180 + 4sin50*10 = 30.64 m/s

v = √22.71² + 30.64² = 38.14 m/s --> speed

I was thinking...should I use 130° instead of 50°, since the acceleration will be in the same direction as the velocity?

 

[gneill]

Ok. I calculated the velocity like so:

v_x = 3cos180 + 4cos50*10 = 22.71 m/s
v_y = 3sin180 + 4sin50*10 = 30.64 m/s

v = √22.71² + 30.64² = 38.14 m/s --> speed

That looks fine.

I was thinking...should I use 130° instead of 50°, since the acceleration will be in the same direction as the velocity?

No, the change in velocity will be in the same direction as the acceleration vector, but the final velocity will have its own angle (with respect to the x-axis). Work out the angle of the velocity vector from your calculated components of the velocity vector.