# Frequently Made Errors in Mechanics: Kinematics

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Kinematics is the subset of dynamics that only concerns itself with time, displacement, velocity and acceleration. A problem is a kinematics problem if it involves those quantities, but consideration of masses, forces, and work is unnecessary.

### 1. Scalars and Vectors

“A scalar only has magnitude while a vector has magnitude and direction”

X “A scalar is a one-dimensional physical quantity, i.e. one that can be described by a single real number (sometimes signed, …”

“A scalar is a one-dimensional quantity, i.e. one that can be described by a single real number (maybe signed) or a complex number…”

A magnitude is unsigned.  For a real number, the magnitude is the absolute value.

In mathematics, all real numbers and complex numbers are scalars, so a scalar can be negative, or might need two real numbers to express it.

In physics, some scalar quantities can definitely be signed (time, charge, potential).  Others (mass) cannot so far as we know, but even then it is not intrinsic in the definition.

### 2. Distance, Area*, Volume and Speed

Most authorities define these as magnitudes:

“The scalar absolute value (magnitude) of velocity is called speed”

Arguably, this is unfortunate.  It makes it meaningless to discuss relative speed, etc.  It also makes this question ill-defined:

“A car travels on a straight road at speed u for time t and at speed u’ for time t’.  How far is it from its starting point?”

According to the ‘speed is a magnitude’ view, we cannot tell whether the car was reversing for one stage.

Some use the term pseudoscalar for a one-dimensional signed quantity.

For some purposes, it may be more convenient to define your variables as signed scalars, or, equivalently, as one-dimensional vectors.

For magnitudes, the operations of addition and subtraction are not associative.  If I’m travelling at 10kph I can speed up by 5kph then slow down by 12kph, but not in the other order.

*Note: An exception is the area element vector in a surface integral.  Something similar may apply to volumes in higher dimensions.

### 3. Distance versus Displacement

“A runner takes time t to run half way around a circular track radius R.  What is the magnitude of the runner’s average velocity?”

X Distance = ##\pi R##, answer = ##\frac{\pi R}{t}##

That would have been correct had the question asked for the average speed.

Velocity is a vector, and the average velocity is given by ##\hat {\vec v}=\frac{\Delta \vec r}{\Delta t}##, where ##\Delta \vec r## is the displacement vector.  In the present case, the runner ran a straight-line distance 2R, so that is the magnitude of the displacement vector:

Displacement = 2R, Average velocity =##|\hat {\vec v}|=\frac{|\Delta \vec r|}{|\Delta t|} = \frac{2R}t##

Displacement always means a vector, so its magnitude is the straight line distance between start and end points.
Distance can mean the magnitude of the displacement, but it can also mean the length along a curve.  Which is intended should be clear from the context.

### 4. Average versus Instantaneous

“A ball takes time t to roll down a ramp length D.  How fast is it moving at the bottom?”

X Speed = distance over time = ##\frac Dt##

Total distance over total time gives the average speed.  The instantaneous speed at a given point may be different.

Assuming constant acceleration down the ramp, the final speed can be obtained from the SUVAT equations.

### 5. Average of given speeds

“A cyclist cycles up a hill at speed u, then back down at speed 3u.  What is the cyclist’s average speed for the round trip?”

X ##\frac{u+3u}2 = 2u##

This is a form of the well-known solecism of taking an average of an average.

Suppose the length of the hill is D.  The time to climb the hill is ##\frac Du##, and the return time is ##\frac D{3u}##, for a total of ##\frac Du\left(1+\frac 13\right)= \frac {4D}{3u}##.  The average speed is total distance / total time =

##\frac{2D}{\frac {4D}{3u}} = \frac 32 u##.

### 6.Velocity versus Acceleration

X “If a particle is stationary it is not accelerating”

It depends what is meant by stationary here.  If stationary for an extended time then clearly it is not accelerating, but if it is only stationary instantaneously (i.e. zero velocity at that instant) then it is accelerating.  A stone thrown straight up has zero velocity at its highest point, but its acceleration is g throughout.

### 7. SUVAT

The “SUVAT” equations only apply when the acceleration is known to be constant.  The acronym refers to the five standard variable names the equations use:

Name Meaning
s Displacement
u Initial velocity
v Final velocity
a Acceleration
t Time

If any three of these are known, the remaining two can be determined.  Thus, there are five SUVAT equations, each omitting one variable:

Omitted Equation
s ##v = u+at##
u ##s = vt-\frac 12 at^2##
v ##s = ut+\frac 12 at^2##
a ##s = \frac{u+v}2 t##
t ##v^2-u^2=2as##

To apply SUVAT equations to a problem:

1. Check that acceleration can be assumed constant.
2. Identify the variable to be determined.
3. Identify three other variables whose values are known, or which will feature in other equations related to the problem.
4. Select the equation involving those four variables.

Example: “A ball is thrown from ground level at angle ##\theta## above the horizontal at speed v.  How far away does it land?”

1. The acceleration is constant in both the vertical direction (g) and the horizontal direction (0).
2. The horizontal displacement is to be determined.
1. In the horizontal direction, only the acceleration and initial velocity are known.
2. In the vertical direction, the acceleration, initial velocity and displacement are known.
3. The time taken is common to the two directions.
1. Find t from u, a, s in the vertical direction
2. Find s from u, a, t in the horizontal direction.

X “The final vertical velocity is zero because it hits the ground”

Its vertical velocity will be zero after hitting the ground, but the process of hitting the ground involves a large upward acceleration.  Applying a SUVAT equation across the period from before to after impact is therefore not valid.  The ‘final’ velocity here means at the instant before impact.

### 8. Relationship between Kinematics Equations and Dynamics equations

Two of the equations in kinematics are just dynamics equations with mass cancelled out.

SUVAT General Kinematics Dynamics
##v^2-u^2=2as## ##v^2=u^2+2\int \vec a.\vec{ds}## ##\Delta KE = \int\vec F.\vec{ds}##
##v = u+at## ##\vec v = \vec u+\int \vec a.dt## ##\Delta \vec p =\int \vec F.dt##

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10 replies
1. robphy says:

Probably the most frequently made error in kinematics isincorrectly thinking "velocity is distance over time" or, slightly-better-but-still-bad, "velocity is displacement over time".Unfortunately, this probably arises from "a common-sense formula" usually encountered via the "average-velocity formula" [which is puzzlingly used to derive the [instantaneous-]velocity formula]—puzzling because a certain-average of a quantity is used to derive the quantity itself.When misused, the average-velocity of a trip doesn't really help you get the velocity at (say) the endpoint. If you happen to accelerate uniformly from rest, using "displacement/time" gets you an answer which is "only off by a factor of 2"..

• haruspex says:

2. DocZaius says:

For Part 3., your final calculation should take the total distance as 2D, not D. Which would make the average speed 3u/2

This Para: Its vertical velocity will be zero after hitting the ground, but the process of hitting the ground involves a large upward acceleration.  Applying a SUVAT equation across the period from before to after impact is therefore not valid.  The ‘final’ velocity here means at the instant before impact.is not clear.The final vertical velocity may not be zero after hitting the ground. It is zero at the instant of hitting the ground, but not zero before and after hitting the ground.Could you elaborate on the sentence: 'Applying a SUVAT equation across the period from before to after impact is therefore not valid.'Radhakrishnamurty

4. robphy says:

Part of the problem with the typical textbook introduction of “average velocity” is that it often doesn’t clarify that
it’s a time-weighted average of velocities (which they usually have not yet defined).
So, often the student is often left incorrectly assuming that it’s a straight-average of velocities.

For a trip with three piecewise-constant-velocity-legs
##
begin{align*}
vec v_{avg}
&equiv frac{vec v_1Delta t_1+vec v_2Delta t_2+vec v_3Delta t_3}{Delta t_1+Delta t_2+Delta t_3} \
&=
frac{Delta vec x_1+Deltavec x_2+Delta vec x_3}{Delta t_1+Delta t_2+Delta t_3}\
&=
frac{Delta vec x_{total}}{Delta t_{total}}\
end{align*}
##

What is usually also missing is an interpretation of the “average velocity”.
I use the following…
Can the trip from start position to end position over the same interval of time be done with a constant velocity
(rather than a varying one)? Yes, use the average-velocity over that interval.

Only when the velocity is constant over the interval
will the average-velocity equal the velocity….
that’s the only time one can use “[constant] velocity=displacement/time”.

For uniform acceleration ,
##
begin{align*}
vec v_{avg}
&equiv
frac{Delta vec x_{total}}{Delta t_{total}}\
&stackrel{scriptsizerm const a }{=}frac{frac{1}{2}vec a(Delta t)^2 +vec v_iDelta t}{Delta t}\
&stackrel{scriptsizerm const a }{=}frac{1}{2}vec a(Delta t) +vec v_i\
&stackrel{scriptsizerm const a }{=}frac{1}{2}(vec v_f -vec v_i) +vec v_i \
&stackrel{scriptsizerm const a }{=} frac{1}{2}(vec v_f+ vec v_i)
end{align*}
##
which looks like a straight-average of the velocities of two piecewise-constant-velocity-legs,
but it’s really the straight-average of the starting and ending velocities.

(If this rendered poorly, try reading it here:

5. haruspex says:

For Part 3., your final calculation should take the total distance as 2D, not D. Which would make the average speed 3u/2

Thanks!