5-22-98
In 1 dimension, we wrote down some general equations relating velocity to displacement, and relating acceleration to the change in velocity, and we also wrote down the four equations that apply in the special case where the acceleration is constant. We're going to do the same thing in 2 dimensions, and the equations will look similar; this shouldn't be surprising because, as we will see, a two (or three) dimensional problem can always be broken down into two (or three) 1-dimensional problems.
When we're dealing with more than 1 dimension (and we'll focus on 2D, but we could use these same equations for 3D), the position is represented by the vector r. The velocity will still be represented by v and the acceleration by a. In general, the average velocity will be given by:
The instantaneous velocity is given by a similar formula, with the condition that a very small time interval is used to measure the displacement.
A similar formula gives the average acceleration:
Again, the instantaneous acceleration is found by measuring the change in velocity over a small time interval.
When the acceleration is constant, we can write out four equations relating the displacement, initial velocity, velocity, acceleration, and time for each dimension. Like the 1D equations, these apply under the following conditions:
If we focus on two dimensions, we get four equations for the x direction and four more for the y direction. These four x equations relate the x-components, while the four y equations relate the y-components.
One thing to notice is that the t, time, is the only thing that doesn't involve an x or a y. This is because everything else is a vector (or a component of a vector, if you'd rather look at it that way), but time is a scalar. Time is the one thing that can be used directly in both the x and y equations; everything else (displacement, velocity, and acceleration) has to be split into components.
Something that probably can't be emphasized enough is that even though an object may travel in a two-dimensional path (often following a parabola, in the standard case of an object moving under the influence of gravity), the motion can always be reduced to two independent one-dimensional motions. The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction.
One good example of this is the case of two objects (e.g., baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown.
We will generally neglect the effect of air resistance in most of the problems we do. In some cases that's just fine. In other cases it's not so fine. A feather and a brick dropped at the same time from the same height will not reach the ground at the same time, for example. This has nothing to do with the weight of the feather compared to the brick. It's simply air resistance; if we took away all the air and dropped the feather and brick, they would hit the ground at exactly the same time.
So, remember that we're often analyzing ideal cases, especially this early in the semester. In reality, things might be a little different because of factors we're neglecting at the moment.
Probably the simplest way to see how to apply the constant acceleration equations to a 2-dimensional problem is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below. You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20° above the horizontal.
(a) How long does it take to reach the ground?
(b) How far is it from the base of the cliff to the point of impact?
It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram.
Then set up a table to keep track of everything you know. Note that you don't have to choose the positive directions the same way, and the origin the same place, as is given here. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes).
Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation:
This gives vy2 = 2.8732 + 2 (-9.8) (-150) = 2948.3 m2 / s2 . Taking the square root gives: vy = +/- 54.30 m/s.
Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground.
Now we can use another equation to solve for time:
So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s.
We can use this time for part (b), to get the distance travelled in the x-direction during the course of its flight. The best equation to use is:
So, from the base of the cliff to the point of impact is 46.0 m.
At some point in its flight, the ball in the example above returned to the level of the top of the cliff (you threw it from the top of the cliff, it went up, and on its way down it passed through a point the same height off the ground as the top of the cliff). What was the ball's velocity when it passed this height? Its speed will be the same as the initial speed, 8.40 m/s, and its angle will be the same as the launch angle, only measured below the horizontal.
This is not just true of the initial height. At every height the ball passes through on the way up, there is a mirror-image point (at the same height, with the same speed, and the same angle, just down rather than up) on the downward part of the path.
Most people find relative velocity to be a relatively difficult concept. In one dimension, however, it's reasonably straight-forward. Let's say you're walking along a road, at 8 km/hr, heading west. A train track runs parallel to the road and a train is passing by, travelling at 40 km/hr west. There is also a car driving by on the road, going 30 km/hr east. How fast is the train travelling relative to you? How fast is the car travelling relative to you? And how fast is the train travelling relative to the car?
One way to look at it is this: in an hour, the train will be 40 km west of where you are now, but you will be 8 km west, so the train will be 32 km further west than you in an hour. Relative to you, then, the train has a velocity of 32 km/hr west. Similarly, relative to the train, you have a velocity of 32 km/hr east.
Using a subscript y for you, t for the train, and g for the ground, we can say this:
the velocity of you relative to the ground = 
= 8 km/hr west
the velocity of the train relative to the ground = 
= 40 km/hr west
Note that if you flip the order of the subscripts, to get the velocity of the ground relative to you, for example, you get an equal and opposite vector. You can write this equal and opposite vector by flipping the sign, or by reversing the direction, like this:
the velocity of the ground relative to you = 
= -8 km/hr west = 8 km/hr east
The velocity of the train relative to you, 
, can be found by adding vectors appropriately. Note the order of the subscripts in this equation:
Rearranging this gives:
A similar argument can be used to show that the velocity of the car relative to you is 38 km/hr east.
the velocity of you relative to the ground = = 8 km/hr west = -8 km/hr east
the velocity of the car relative to the ground = 
= 30 km/hr east
The velocity of the train relative to the car is 70 km/hr west, and the velocity of the car relative to the train is 70 km/hr east.
In two dimensions, the relative velocity equations look identical to the way they look in one dimension. The main difference is that it's harder to add and subtract the vectors, because you have to use components. Let's change the 1D example to 2D. The train still moves at 40 km/hr west, but the car turns on to a road going 40° south of east, and travels at 30 km/hr. What is the velocity of the car relative to the train now?
The relative velocity equation for this situation looks like this:
The corresponding vector diagram looks like this:
Because this is a 2-D situation, we have to write this as two separate equations, one for the x-components (east-west) and one for the y-components (north-south):
Now we have to figure out what the x and y components are for these vectors. The train doesn't have a y-component, because it is travelling west. So:
The car has both an x and y component:
Plugging these in to the x and y equations gives:
Combining these two components into the vector gives a magnitude of:
at an angle given by the inverse tangent of 19.3 / 63.0, which is 17°. So, the velocity of the car relative to the train is 66 km/hr, 17° south of east.