SOLVING KINEMATICS PROBLEMS
IN A NUTSHELL
Kinematics is the part of the course where you can still sort of get away with taking equations that are given to you, and substituting ("plugging in") values in these equations. That won't last, and even in kinematics, you need to understand the physical situation in order to know what to substitute.
The equations that describe the
motion of an object that has constant acceleration ax are:
2) vx = vx0 + ax t
These two are the most fundamental. They come from integrating once (vx ) and twice (x) with respect to time a constant acceleration ax.
3) (vx )2 - (vx0)2 = 2 ax (x - x0)
This equation is derived from the above two.
4) x - x0 = 1/2 ( vx0 + vx) t
This is because when the acceleration is constant, 1/2 ( vx0 + vx) is the average speed; so this equation means "The distance traveled is equal to the average speed multiplied by the time of travel".
In these equations t is the time duration of interest, but some books use it to mean a moment of time, and then use t - t0 instead of t in the equations above (then t0 represents the instant, i.e. the moment, at the start of the time interval of interest and t the instant at the end of it).
You might think: "So, that's ALL the equations I need, right?".... WRONG!
Unless the problem is simple, you will need this group of equations several times:
a) if the object(s) changes acceleration, you have to divide the problem into different time intervals over which the acceleration remains the same, and then you need as many groups of these equations as there are different time intervals: i.e.the object will have different accelerations at different times, that must be constant throughout each of these time durations, and for each of these you will have a set of the equations above. For instance, for equation 1:
x1 = x0 + v0 t1 + 1/2 a1 t12 : the object has acceleration a1 for a length of time t1.
and x2 = x1 + v1 t2 + 1/2 a2 t22: just afterwards (so that the initial velocity during that time is v1the final velocity for the time interval), the object has acceleration a2 for a length of time t2.
b) if there are several objects (for instance, a car and a truck, two trains, etc...), then you need these equations for each of these objects. So you need to use subscripts to tell the objects apart. For instance, t for truck and c for car, so that for the first equation you would get:
xt = xt0 + vtx0 t + 1/2 atx tt2 and xc = xc0 + vcx0 t + 1/2 acx tc2
Usually however these objects will move along just one direction, i.e. in one dimension so you can drop the subscript x:
xt = xt0 + vt0 t + 1/2 at tt2 and xc = xc0 + vc0 t + 1/2 ac tc2
When you realize that two quantities of the same kind have the same value, you can drop the subscript. For instance, if you know that tt = tc, you can just write t.
c) if the object is moving in 2 dimensions, such as in projectile motion for instance, you need another group of these equations with y instead of x.
Projectile motion mean that the object is moving only under the influence of gravity (ex: something that is thrown, a vehicle having driven off a cliff; but not something that flies).
When an object is undergoing "projectile motion", the following is true:
- all its acceleration is vertical and downwards, and its magnitude is g (near the Earth). So:
ax = 0 m/s2and ay = - g = - 9.8 m/s2.
- Also, because ax = 0, the horizontal component of the object's velocity is constant:
vx = v0 = constant
WHAT TO DO WITH THESE EQUATIONS:
- Simplify them as much as you can: figure out which variables happen to be 0 in the case you are dealing with, and "kill off" the corresponding terms (a term is an expression that is either added or subtracted to another, say: vt0 t for example, or 1/2 at tt2).
Some of the values will be given to you in words, not as a number (usually when a value is 0!).
So remember:
- Identify in the equations which other variables you know the value of and which you don't.
- use the equation(s) that contain(s) the variable you are looking for, and, if possible none of the variables that you don't know. You will usually end up with several equations with several unknowns, and you will have to solve this system (by whatever means necessary! Substitution is usually a favorite).
IF THINGS GET TOO COMPLICATED:
Usually a good thing to do then is this: for each time interval, each direction and each object, list all relevant quantities, indicating their numerical values whenever you know them. For example for the car and truck discussed above, you would list:
1) x = x0
+ vx0 t + 1/2 ax t2
2) vx = vx0 + ax t
3) (vx )2 - (vx0)2 = 2 ax (x - x0)
4) x - x0 = 1/2 ( vx0 + vx) t
In these equations t is the time duration of interest, but some books use it to mean a moment of time, and then use t - t0 instead of t in the equations above (then t0 represents the instant, i.e. the moment, at the start of the time interval of interest and t the instant at the end of it).
You might think: "So, that's ALL the equations I need, right?".... WRONG!
Unless the problem is simple, you will need this group of equations several times:
a) if the object(s) changes acceleration, you have to divide the problem into different time intervals over which the acceleration remains the same, and then you need as many groups of these equations as there are different time intervals: i.e.the object will have different accelerations at different times, that must be constant throughout each of these time durations, and for each of these you will have a set of the equations above. For instance, for equation 1:
x1 = x0 + v0 t1 + 1/2 a1 t12 : the object has acceleration a1 for a length of time t1.
and x2 = x1 + v1 t2 + 1/2 a2 t22: just afterwards (so that the initial velocity during that time is v1the final velocity for the time interval), the object has acceleration a2 for a length of time t2.
b) if there are several objects (for instance, a car and a truck, two trains, etc...), then you need these equations for each of these objects. So you need to use subscripts to tell the objects apart. For instance, t for truck and c for car, so that for the first equation you would get:
xt = xt0 + vtx0 t + 1/2 atx tt2 and xc = xc0 + vcx0 t + 1/2 acx tc2
Usually however these objects will move along just one direction, i.e. in one dimension so you can drop the subscript x:
xt = xt0 + vt0 t + 1/2 at tt2 and xc = xc0 + vc0 t + 1/2 ac tc2
When you realize that two quantities of the same kind have the same value, you can drop the subscript. For instance, if you know that tt = tc, you can just write t.
c) if the object is moving in 2 dimensions, such as in projectile motion for instance, you need another group of these equations with y instead of x.
Projectile motion mean that the object is moving only under the influence of gravity (ex: something that is thrown, a vehicle having driven off a cliff; but not something that flies).
When an object is undergoing "projectile motion", the following is true:
- all its acceleration is vertical and downwards, and its magnitude is g (near the Earth). So:
ax = 0 m/s2and ay = - g = - 9.8 m/s2.
- Also, because ax = 0, the horizontal component of the object's velocity is constant:
vx = v0 = constant
WHAT TO DO WITH THESE EQUATIONS:
- Simplify them as much as you can: figure out which variables happen to be 0 in the case you are dealing with, and "kill off" the corresponding terms (a term is an expression that is either added or subtracted to another, say: vt0 t for example, or 1/2 at tt2).
Some of the values will be given to you in words, not as a number (usually when a value is 0!).
So remember:
- "starts from rest" means that the object's initial velocity is 0 m/s .
- "stops" means that the object's "final" velocity is 0 m/s (i.e. its velocity at the end of the time interval in question).
- in projectile motion, ax = 0 m/s2 , ay = - g = - 9.8 m/s2 , vx = v0 = constant.
- at the highest point of its path, a projectile has the vertical component of its velocity equal to 0 (for an instant, it is moving neither up nor down). So at that moment / position, vy = 0 m/s.
- Identify in the equations which other variables you know the value of and which you don't.
- use the equation(s) that contain(s) the variable you are looking for, and, if possible none of the variables that you don't know. You will usually end up with several equations with several unknowns, and you will have to solve this system (by whatever means necessary! Substitution is usually a favorite).
IF THINGS GET TOO COMPLICATED:
Usually a good thing to do then is this: for each time interval, each direction and each object, list all relevant quantities, indicating their numerical values whenever you know them. For example for the car and truck discussed above, you would list:
ac: at:
xc0: xc : xt0: xt :
vc0: vc : vt0: vt :
tc : tt:
where I have assumed that both move in the same, one dimension.
If you had an object changing acceleration, you would have something similar with subscripts for the different intervals of time rather than for the different objects. And if you were dealing with an object moving in two dimensions you'd have instead:
ax: ay:
x0: x : y0: y :
vx0: vx : vy0: vy :
t :
You would indicate each of the values that you are given in the text of the problem, or that you can figure out using physical intuition (and leave the rest blank for now!)
Out of the various kinematics
equations, use the equation(s) that contain(s) the variable you are looking for,
and, if possible none of the variables that you don't know. For instance, say you want the distance traveled and the only thing you don't know is the final velocity, use equation 1.
Note: this discussion refers to linear motion, but there are analogous equations for rotation, and
the same technique applies there (only you would be dealing with only one
direction).
