On the time interval from t = 5 seconds to t = 10 seconds, the velocity changes from 15
s
m
to 35
. Thus, on that time interval the acceleration is given by:
s
m
m
35
−15
∆v
v − v
f −
m
i
s
s
a =
=
=
= 4
2
∆ t
t −
f
t
10 s − 5 s
s
i
m
On the time interval from t = 10 seconds to t = 20 seconds, the velocity changes from 35
s
m
to 0
. Thus, on that time interval the acceleration is given by:
s
m
m
0
− 35
∆v
v − v
f −
m
i
s
s
a =
=
=
= 3
− 5
. 2
∆ t
t −
f
t
20 s − 10 s
s
i
51
Chapter 10 Constant Acceleration Problems in Two Dimensions
10 Constant Acceleration Problems in Two Dimensions
In solving problems involving constant acceleration in two dimensions, the most
common mistake is probably mixing the x and y motion. One should do an
analysis of the x motion and a separate analysis of the y motion. The only
variable common to both the x and the y motion is the time. Note that if the initial
velocity is in a direction that is along neither axis, one must first break up the
initial velocity into its components.
In the last few chapters we have considered the motion of a particle that moves along a straight
line with constant acceleration. In such a case, the velocity and the acceleration are always
directed along one and the same line, the line on which the particle moves. Here we continue to
restrict ourselves to cases involving constant acceleration (constant in both magnitude and
direction) but lift the restriction that the velocity and the acceleration be directed along one and
the same line. If the velocity of the particle at time zero is not collinear with the acceleration,
then the velocity will never be collinear with the acceleration and the particle will move along a
curved path. The curved path will be confined to the plane that contains both the initial velocity
vector and the acceleration vector, and in that plane, the trajectory will be a parabola. (The
trajectory is just the path of the particle.)
You are going to be responsible for dealing with two classes of problems involving constant
acceleration in two dimensions:
(1) Problems involving the motion of a single particle.
(2) Collision Type II problems in two dimensions
We use sample problems to illustrate the concepts that you must understand in order to solve
two-dimensional constant acceleration problems.
Example 10-1
A horizontal square of edge length 1.20 m is situated on a Cartesian coordinate
system such that one corner of the square is at the origin and the corner opposite
that corner is at (1.20 m, 1.20 m). A particle is at the origin. The particle has an
initial velocity of 2.20 m/s directed toward the corner of the square at
(1.20 m, 1.20 m) and has a constant acceleration of 4.87 m/s2 in the +x direction.
Where does the particle hit the perimeter of the square?
52
Chapter 10 Constant Acceleration Problems in Two Dimensions
Solution and Discussion
Let’s start with a diagram.
y
(1.20 m, 1.20 m)
v
o
a
x
Now let’s make some conceptual observations on the motion of the particle. Recall that the
square is horizontal so we are looking down on it from above. It is clear that the particle hits the
right side of the square because: It starts out with a velocity directed toward the far right corner.
That initial velocity has an x component and a y component. The y component never changes
because there is no acceleration in the y direction. The x component, however, continually
increases. The particle is going rightward faster and faster. Thus, it will take less time to get to
the right side of the square then it would without the acceleration and the particle will get to the
right side of the square before it has time to get to the far side.
An important aside on the trajectory (path) of the particle: Consider an ordinary checker
on a huge square checkerboard with squares of ordinary size (just a lot more of them then
you find on a standard checkerboard). Suppose you start with the checker on the extreme
left square of the end of the board nearest you (square 1) and every second, you move the
checker right one square and forward one square. This would correspond to the checker
moving toward the far right corner at constant velocity. Indeed you would be moving the
checker along the diagonal. Now let’s throw in some acceleration. Return the checker to
square 1 and start moving it again. This time, each time you move the checker forward,
you move it rightward one more square than you did on the previous move. So first you
move it forward one square and rightward one square. Then you move it forward another
square but rightward two more squares. Then forward one square and rightward three
squares. And so on. With each passing second, the rightward move gets bigger. (That’s
what we mean when we say the rightward velocity is continually increasing.) So what
would the path of the checker look like? Let’s draw a picture.
53
Chapter 10 Constant Acceleration Problems in Two Dimensions
As you can see, the checker moves on a curved path. Similarly, the path of the particle in
the problem at hand is curved.
Now back to the problem at hand. The way to attack these two-dimensional constant
acceleration problems is to treat the x motion and the y motion separately. The difficulty with
that, in the case at hand, is that the initial velocity is neither along x nor along y but is indeed a
mixture of both x motion and y motion. What we have to do is to separate it out into its x and y
components. Let’s proceed with that. Note that, by inspection, the angle that the velocity vector
makes with the x axis is 45.0° .
54
Chapter 10 Constant Acceleration Problems in Two Dimensions
y
vo =
2.20 m/s
v
oy
θ = 45°
vox
x
v
By inspection (because the angle is 45.0°):
ox
cosθ = v
o
v = v
v
= v
cosθ
oy
ox
ox
o
So:
m
m
o
v =
v = .
2 20
.
cos 45 0
.
1 556
oy
ox
s
s
m
v = 1.556
ox
s
Now we are ready to attack the x motion and the y motion separately. Before we do, let’s
consider our plan of attack. We have established, by means of conceptual reasoning, that the
particle will hit the right side of the square. This means that we already have the answer to half
of the question “Where does the particle hit the perimeter of the square?” It hits it at x = 1.20 m
and y = ? . All we have to do is to find out the value of y. We have established that it is the
x motion that determines the time it takes for the particle to hit the perimeter of the square. It
hits the perimeter of the square at that instant in time when x achieves the value of 1.20 m. So
our plan of attack is to use one or more of the x-motion constant acceleration equations to
determine the time at which the particle hits the perimeter of the square and to plug that time into
the appropriate y-motion constant acceleration equation to get the value of y at which the particle
hits the side of the square. Let’s go for it.
55
Chapter 10 Constant Acceleration Problems in Two Dimensions
x motion
We start with the equation that relates position and time:
0
1
2
x = x + v t +
a t (We need to find the time that makes x = 1.20 m.)
o
ox
x
2
The x component of the acceleration is the total acceleration, that is ax = a. Thus,
1
2
x = v t +
a t
ox
2
Recognizing that we are dealing with a quadratic equation we get it in the standard form of the
quadratic equation.
1
2
a t + v t − x = 0
2
ox
Now we apply the quadratic formula:
2
1
−v ± v − 4
a ( x)
ox
ox
−
2
t =
1
2 ax
2
2
−v ± v + 2a x
ox
ox
t =
ax
Substituting values with units (and, in this step, doing no evaluation) we obtain:
2
m
m
m
−1 556
.
± 1 556
.
+ 2 4 87
.
1.
m
20
2
s
s
s
t =
m
4.87 2
s
Evaluating this expression yields:
t = 0.4518 s, and t = −1.091 s.
We are solving for a future time so we eliminate the negative result on the grounds that it is a
time in the past. We have found that the particle arrives at the right side of the square at time
t = 0.4518 s. Now the question is, “What is the value of y at that time?”
56
Chapter 10 Constant Acceleration Problems in Two Dimensions
y-motion
Again we turn to the constant acceleration equation relating position to time, this time writing it
in terms of the y variables:
0
0
1
2
y = y +v t +
a t
o
oy
y
2
We note that y0 is zero because the particle is at the origin at time 0 and ay is zero because the
acceleration is in the +x direction meaning it has no y component. Rewriting this:
y =v t
oy
Substituting values with units,
m
y = 1.556
(0.
s
4518 )
s
evaluating, and rounding to three significant figures yields:
y = 0.703 m.
Thus, the particle hits the perimeter of the square at
(1.20 m, 0.703 m)
Next, let’s consider a 2-D Collision Type II problem. Solving a typical 2-D Collision Type II
problem involves finding the trajectory of one of the particles, finding when the other particle
crosses that trajectory, and establishing where the first particle is when the second particle
crosses that trajectory. If the first particle is at the point on its own trajectory where the second
particle crosses that trajectory then there is a collision. In the case of objects rather than
particles, one often has to do some further reasoning to solve a 2-D Collision Type II problem.
Such reasoning is illustrated in the following example involving a rocket.
57
Chapter 10 Constant Acceleration Problems in Two Dimensions
Example 10-2
The positions of a particle and a thin (treat it as being as thin as a line) rocket of
length 0.280 m are specified by means of Cartesian coordinates. At time 0 the
particle is at the origin and is moving on a horizontal surface at 23.0 m/s at 51.0°. It
has a constant acceleration of 2.43 m/s2 in the +y direction. At time 0 the rocket is at
rest and it extends from (−.280 m, 50.0 m) to (0, 50.0 m), but it has a constant
acceleration in the +x direction. What must the acceleration of the rocket be in order
for the particle to hit the rocket?
Solution
Based on the description of the motion, the rocket travels on the horizontal surface along the line
y = 50.0 m. Let’s figure out where and when the particle crosses this line. Then we’ll calculate
the acceleration that the rocket must have in order for the nose of the rocket to be at that point at
that time and repeat for the tail of the rocket. Finally, we’ll quote our answer as being any
acceleration in between those two values.
When and where does the particle cross the line y = 50.0 m?
We need to treat the particle’s x motion and the y motion separately. Let’s start by breaking up
the initial velocity of the particle into its x and y components.
y
vo =
23.0 m/s
voy
θ = 51.0°
v
x
ox
v
v
ox
cosθ =
oy
v
sinθ =
v
o
o
v = v cosθ
v = v sinθ
ox
o
oy
o
m
o
v = 23 0
.
cos 51 0
.
m
o
v
ox
= 23 0
.
sin 51.0
s
oy
s
m
v = 14.47
m
ox
v = 17.87
s
oy
s
58
Chapter 10 Constant Acceleration Problems in Two Dimensions
Now in this case, it is the y motion that determines when the particle crosses the trajectory of the
rocket because it does so when y = 50.0 m. So let’s address the y motion first.
y motion of the particle
0
1
2
y = y +v t + a t
o
oy
y
2
Note that we can’t just assume that we can cross out y o but in this case the time zero position of
the particle was given as (0, 0) meaning that y o is indeed zero for the case at hand. Now we
solve for t:
1
2
y =v t +
a t
oy
y
2
1
2
a t +v t − y = 0
2 y
oy
2
1
−v ± v − 4
a
( y)
oy
oy
y −
2
t =
1
2 ay
2
2
−v
± v
+ 2 a y
oy
oy
y
t =
ay
2
m
m
m
−17 87
.
± 17.87 + 2 2 43
.
m
0
.
50
2
s
s
s
t =
m
2 43
.
2
s
t = 2.405 s, and t = −17.11 s.
Again, we throw out the negative solution because it represents an instant in the past and we
want a future instant.
Now we turn to the x motion to determine where the particle crosses the trajectory of the rocket.
59
Chapter 10 Constant Acceleration Problems in Two Dimensions
x motion of the particle
Again we turn to the constant acceleration equation relating position to time, this time writing it
in terms of the x variables:
0
(because the particle starts at the origin)
0 (because the acceleration is in the y direction)
1
2
x = x +v t +
a t
o
ox
x
2
x =v t
ox
m
x = 14.47
(2.
s
405 )
s
x = 34. 80 m.
So the particle crosses the rocket’s path at (34.80 m, 50.0 m) at time t = 2.450 s. Let’s calculate
the acceleration that the rocket would have to have in order for the nose of the rocket to be there
at that instant. The rocket has x motion only. It is always on the line y = 50.0 m.
Motion of the Nose of the Rocket
0
0
1
2
x′ = x′ +v ′ t +
a′t
n
on
oxn
n
2
where we use the subscript n for “nose” and a prime to indicate “rocket.” We have crossed out
x′ because the nose of the rocket is at (0, 50.0 m) at time zero, and we have crossed out v ′
on
oxn
because the rocket is at rest at time zero.
1
2
x′ =
a′t
n
n
2
Solving for a′ yields:
n
2x′n
a′ =
n
2
t
Now we just have to evaluate this expression at t = 2.405 s, the instant when the particle crosses
the trajectory of the rocket, and at x′ = x = 34.80 m, the value of x at which the particle crosses
n
the trajectory of the rocket.
34
(
2
.80 m)
a′ =
n
2
(2 405
.
s)
m
a′ = 12.0
n
2
s
It should be emphasized that the n for “nose” is not there to imply that the nose of the rocket has
a different acceleration than the tail; rather; the whole rocket must have the acceleration
60
Chapter 10 Constant Acceleration Problems in Two Dimensions
m
a′ = 12 0
.
in order for the particle to hit the rocket in the nose. Now let’s find the acceleration
n
2
s
a′ that the entire rocket must have in order for the particle to hit the rocket in the tail.
t
Motion of the Tail of the Rocket
0
1
2
x′ = x′ +v ′ t +
a′ t
t
ot
oxt