How to Handle Velocity as a Vector
Velocity is a vector, and as such, it has a magnitude and a direction associated with it. Suppose you’re in a car traveling east at 88 meters/second when you begin to accelerate north at 5.0 meters/second^{2} for 10.0 seconds. What is your final speed?
You may think you can use this equation to figure out the answer:
v_{f}_{ }= v_{o} + a x t
But that’s not a vector equation; the quantities here are called scalars (the magnitude of a vector is a scalar). This is a scalar equation, and it’s not appropriate to use here because the acceleration and the initial speed aren’t in the same direction. In fact, speed itself is a scalar, so you have to think in terms not of speed but of velocity.
Here’s the same equation as a vector equation:
v_{f} = v_{o} + a x t
Note that the speeds are now velocities (speed is the magnitude of a velocity vector) and that everything here is a vector except time (which is always a scalar). This change means that the addition you perform in this equation is vector addition, which is what you want because vectors can handle addition in multiple dimensions, not just in a straight line.
Here are the equations of motion, written as vector equations:
v_{f} = v_{o} + a x t
Sample question

You’re in a car traveling east at 88.0 meters/second; then you accelerate north at 5.00 meters/second^{2} for 10.0 seconds. What is your final speed?
The correct answer is 101 meters/second.

Start with this vector equation:
v_{f} = v_{o} + a x t

This equation is simply vector addition, so treat the quantities involved as vectors.
That is, v_{o} = (88, 0) meters/second and a = (0, 5) meters/second^{2}. Here’s what the equation looks like when you plug in the numbers:
v_{f} = (88.0, 0) + (0, 5.00)(10.0)

Do the math:
v_{f} = (88.0, 0) + (0, 5.00)(10.0) = (88.0, 50.0)

You’re asked to find the final speed, which is the magnitude of the velocity. Plug your numbers into the Pythagorean theorem.

You can also find the final direction.
Apply the equation theta = tan^{1}(y/x) to find the angle, which is tan^{–1}(50.0/88.0) = tan^{–1}(0.57) = 29.6 degrees in this case.

Practice questions

You’re going 40.0 meters/second east, and then you accelerate 10.0 meters/second^{2} north for 10.0 seconds. What are the direction and magnitude of your final velocity?

You’re going 44.0 meters/second at 35 degrees, and then you accelerate due west at 4.0 meters/second^{2} for 20.0 seconds. What are the direction and magnitude of your final velocity?

A hockey puck is going 100.0 meters/second at 250 degrees when it’s hit by a hockey stick, which accelerates it at 1.0 x 10^{3} meters/second^{2} at 19 degrees for 0.10 second. What are the direction and magnitude of the puck’s final velocity?

A car is driving along an icy road at 10.0 meters/second at 0 degrees when it skids, accelerating at 15 meters/second^{2} at 63 degrees for 1.0 second. What are the direction and magnitude of the car’s final velocity?
Following are answers to the practice questions:

Magnitude 108 meters/second, angle 68 degrees

Start with this equation: v_{f} = v_{o} + a x t.

Plug in the numbers: v_{f} = (40.0, 0) + (0, 10.0)(10.0) = (40.0, 100.0).

Convert the vector (40.0, 100.0) into magnitude/angle form. Use the equation theta = tan^{–1}(y/x) to find the angle: tan^{–1}(100.0/40.0) = tan^{–1}(2.5) = 68 degrees.

Apply the equation
to find the speed — the magnitude of the velocity, giving you 108 meters/second.


Magnitude 50.7 meters/second, angle 150 degrees

Start with this equation: v_{f} = v_{o} + a x t.

Convert the original velocity into vector component notation.
Use the equation v_{x} = v cos theta to find the x coordinate of the original velocity vector: 44.0 x cos 35 degrees = 36.0.

Use the equation v_{y} = v sin theta to find the y coordinate of the velocity: 44.0 x sin 35 degrees, or 25.2. So the velocity is (36.0, 25.2) in coordinate form.

Perform the vector addition: (36.0, 25.2) + (–4.0, 0)(20.0) = (–44.0, 25.2).

Convert the vector (–44.0, 25.2) into magnitude/angle form.
Use the equation theta = tan^{–1}(y/x) to find the angle: tan^{–1}(25.2/–44.0) = tan^{–1}(0.57) = 150 degrees.

Apply the equation
to find the speed — the magnitude of the velocity, giving you 50.7 meters/second.


Magnitude 86.1 meters/second, angle –46 degrees

Start with this equation: v_{f} = v_{o} + a x t.

Convert the original velocity into vector component notation.
Use the equation v_{x} = v cos theta to find the x coordinate of the original velocity vector: 100.0 x cos 250 degrees = –34.2.

Use the equation v_{y} = v sin theta to find the y coordinate of the velocity: 100.0 x sin 250 degrees, or –94.0.
So the original velocity is (–34.2, –94.0) in coordinate form.

Convert the acceleration into components.
Use the equation a_{x} = a cos theta to find the x coordinate of the acceleration: (1.0 x 10^{3})cos 19 degrees = 946.

Use the equation a_{y} = a sin theta to find the y coordinate of the acceleration: (1.0 x 10^{3})sin 19 degrees, or 325.
So the acceleration is (946, 325) in coordinate form.

Perform the vector addition: (–34.2, –94.0) + (945, 325)(0.1) = (60.3, –61.5).

Convert the vector (60.3, –61.5) into magnitude/angle form.

Use the equation theta = tan^{–1}(y/x) to find the angle: tan^{–1}(61.5/60.3) = tan^{–1}(–1.0) = –46 degrees.

Apply the equation
to find the speed — the magnitude of the velocity, giving you 86.1 meters/second.


Magnitude 21.5 meters/second, angle 39 degrees

Start with this equation: v_{f} = v_{o} + a x t.

Convert the original velocity into vector component notation: (10.0, 0) meters/second.

Convert the acceleration into components. Use the equation a_{x} = a cos theta to find the x coordinate of the acceleration: 15 x cos 63 degrees = 6.8.

Use the equation a_{y} = a sin theta to find the y coordinate of the acceleration: 15 x sin 63 degrees, or 13.4.
So the acceleration is (6.8, 13.4) in coordinate form.

Perform the vector addition: (10.0, 0) + (6.8, 13.4)(1.0) = (16.8, 13.4).

Convert the vector (16.8, 13.4) into magnitude/angle form.
Use the equation theta = tan^{–1}(y/x) to find the angle: tan^{–1}(13.4/16.8) = tan^{–1}(0.79) = 39 degrees.

Apply the equation
to find the magnitude of the velocity, giving you 21.5 meters/second.
