The language of kinematics describes the motion of particles or bodies. We will only consider motion in one dimension (a single straight line), where all measurements are related to the origin, just like the $0$0 value on a number line. The ideas we develop in one dimension can then be extended to two- or three-dimensional motion.
The letter $s$s is often used to represent position, that is, displacement relative to the origin. Just like on a number line, this quantity can be negative. We use $t$t to represent elapsed time from some initial starting point, represented by the equation $t=0$t=0.
We then introduce velocity $v$v as the rate of change of displacement. Like displacement (but unlike the more colloquial "speed"), this quantity can be negative. Let's take some time to define it carefully.
We use the symbol $\Delta$Δ (pronounced "Delta") to indicate "change", so $\Delta s$Δs represents a change in displacement while $\Delta t$Δt represents a change in time. We say the average velocity $\overline{v}$v over that time is
$\overline{v}=\frac{\Delta s}{\Delta t}$v=ΔsΔt.
So a particle travelling $\Delta s=30$Δs=30 m in $\Delta t=10$Δt=10 s has an average velocity of $\overline{v}=\frac{30}{10}=3$v=3010=3 m/s over those $10$10 seconds. This does not mean that the particle travelled at exactly $3$3 m/s for the whole $10$10 seconds, however!
The velocity might be changing during the interval $\Delta t$Δt beginning at $t=0$t=0 and ending at $t=10$t=10, but if we make $\Delta t$Δt smaller and smaller, the average velocity over the shorter interval will become closer and closer to the "real" velocity around that time. Drawing from calculus, we therefore define the instantaneous velocity $v$v to be
$v=\lim_{\Delta t\rightarrow0}\frac{\Delta s}{\Delta t}$v=limΔt→0ΔsΔt.
This means that when displacement $s$s is given as a smoothly varying function of $t$t, the velocity function $v(t)$v(t) is the derivative of the displacement function:
$v(t)=\frac{ds}{dt}$v(t)=dsdt.
If the velocity is constant then it is equal to the average velocity over any time interval, and we can write simply $v=\frac{s}{t}$v=st or, after rearrangement, $s=vt$s=vt.
When velocity varies, we say there is an acceleration $a$a. We define acceleration to be the rate of change of velocity - the derivative of the velocity function with respect to time:
$a(t)=\frac{dv}{dt}$a(t)=dvdt.
Observe that the acceleration function is the second derivative of the displacement function:
$a(t)$a(t) | $=$= | $\frac{dv}{dt}$dvdt |
$=$= | $\frac{d}{dt}\frac{ds}{dt}$ddtdsdt | |
$=$= | $\frac{d^2s}{dt^2}$d2sdt2 |
There are four formulae that are applicable when acceleration is constant over some time interval. This occurs, for example, when an object not too far above the ground is dropped - such an object falls with constant acceleration (as long as we ignore the effect of air resistance).
The formulae connect displacement $s$s, initial velocity $u$u, final velocity $v$v, and constant acceleration $a$a, through the variable of elapsed time $t$t. They can be derived nicely by consideration of a velocity-time graph with a constant gradient, or we can use some ideas from calculus.
Given that $a(t)=\frac{dv}{dt}$a(t)=dvdt with $a(t)$a(t) a constant, say $a$a, we can antidifferentiate both sides:
$\int a\ dt$∫a dt | $=$= | $\int\frac{dv}{dt}\ dt$∫dvdt dt |
$=$= | $\int dv$∫dv | |
$\therefore\ at+C$∴ at+C | $=$= | $v$v |
At $t=0$t=0 we know that $v=u$v=u. This means that the constant of integration $C$C is $u$u, and we have produced the first kinematic formula:
$v=u+at$v=u+at | equation (1) |
Now, velocity, as a function of time, is the derivative of the displacement function. So, we can rewrite equation (1) as $\frac{ds}{dt}=u+at$dsdt=u+at. Once again we antidifferentiate both sides:
$\int\frac{ds}{dt}\ dt$∫dsdt dt | $=$= | $\int u+at\ dt$∫u+at dt |
$\int ds$∫ds | $=$= | $ut+\frac{1}{2}at^2+C$ut+12at2+C |
$s$s | $=$= | $ut+\frac{1}{2}at^2+C$ut+12at2+C |
If the displacement at time $t=0$t=0 is considered to be zero, then the constant $C$C is zero and we produce the second kinematic formula:
$s=ut+\frac{1}{2}at^2$s=ut+12at2 | equation (2) |
We use equations (1) and (2) to obtain the last two. We rewrite equation (1) to make $a$a the subject, obtaining $a=\frac{v-u}{t}$a=v−ut, and substitute this into equation (2). This produces
$s$s | $=$= | $ut+\frac{1}{2}\frac{v-u}{t}t^2$ut+12v−utt2 |
$=$= | $t\left(u+\frac{v-u}{2}\right)$t(u+v−u2) | |
$=$= | $t\frac{u+v}{2}$tu+v2 |
This can be written as
$\frac{s}{t}=\frac{u+v}{2}$st=u+v2,
which, in words, expresses that (when acceleration is constant) the average velocity throughout a time period is simply the average of the initial and final velocities.
If we rewrite equation (1) to make $t$t the subject, obtaining $t=\frac{v-u}{a}$t=v−ua, and substitute this into equation (2), we obtain
$s$s | $=$= | $u\frac{v-u}{a}+\frac{1}{2}a\left(\frac{v-u}{a}\right)^2$uv−ua+12a(v−ua)2 |
$=$= | $\left(\frac{v-u}{a}\right)\left(u+\frac{1}{2}a\frac{v-u}{a}\right)$(v−ua)(u+12av−ua) | |
$=$= | $\left(\frac{v-u}{a}\right)\left(\frac{v+u}{2}\right)$(v−ua)(v+u2) | |
$=$= | $\frac{v^2-u^2}{2a}$v2−u22a |
This is usually written as
$v^2-u^2=2as$v2−u2=2as,
and can be the most useful equation to relate the quantities $v$v, $u$u, $a$a, and $s$s without needing to substitute (or even know!) the value of $t$t.
If a particle or body has initial velocity $u$u, final displacement $s$s, final velocity $v$v, constant acceleration $a$a, all over a length of time $t$t, the following four kinematic equations hold:
By identifying the known quantities in a question, you will be able to solve for the other values by choosing the right kinematic equation.
Solve: A projectile is launched vertically upwards from ground level with an initial velocity of $235$235 m/s. How high does it reach?
Think: We have been given the initial velocity $u=235$u=235, and want to know its displacement $s$s when it stops moving. We can represent "the projectile stops moving" by setting the final velocity to be zero, $v=0$v=0. The acceleration due to gravity $a$a is known, and is approximately $-9.8$−9.8 m/s - the negative sign indicates that the acceleration is directed downwards. The equation that links $u$u, $s$s, $v$v, and $a$a all together is the fourth kinematic equation, $v^2-u^2=2as$v2−u2=2as.
Do: Substituting the given values into the formula $v^2-u^2=2as$v2−u2=2as leaves us with an equation where only $s$s is unknown:
$0^2-235^2=2\times\left(-9.8\right)\times s$02−2352=2×(−9.8)×s.
After performing the arithmetic we can find the correct value for the projectile's displacement at its highest point, $s=\frac{-235^2}{2\times(-9.8)}=2817.6$s=−23522×(−9.8)=2817.6 m.
A particle moves in a straight line and its displacement after $t$t seconds is given by $x=12t-2t^2$x=12t−2t2, where $x$x is its displacement in metres from the starting point. Let $v$v and $a$a represent its velocity and acceleration at time $t$t respectively.
Determine an equation for the velocity $v$v of the particle after $t$t seconds.
After how many seconds $t$t does the particle change its direction of motion?
Write each line of working as an equation.
Plot the graph of displacement v.s. time on the axes below:
Determine the displacement of the particle after $9$9 seconds.
Hence find the total distance that the particle has traveled in the first $9$9 seconds.
A particle moves in a straight line so that after $t$t seconds ($t\ge0$t≥0) its velocity $v$v is given by $v=\frac{3}{1+t}-t+1$v=31+t−t+1 m/s. The displacement of the particle from the origin is given by $x$x metres.
If the particle is initially at the origin, find the displacement as a function of time $t$t.
You may use $C$C to represent the constant of integration if needed.
Solve for the time $t$t at which the particle is stationary.
How far does the particle travel in the first $4$4 seconds? Give your answer to two decimal places.
The displacement of an object is a measure of how far it is from a fixed point. It can be negative or positive depending on which side of the fixed point the object is on.
A particle moves in a straight line and its displacement $x$x cm from a fixed origin point after $t$t seconds is given by the function $x\left(t\right)=\sin t-\sin t\cos t-5t$x(t)=sint−sintcost−5t. Displacement to the left of the origin is negative and displacement to the right of the origin is positive.
Determine the initial displacement of the particle.
The velocity $v\left(t\right)$v(t) is the rate at which displacement changes over time.
Find an equation for $v\left(t\right)$v(t) and express your answer in terms of $\cos t$cost only.
Determine the initial velocity $v\left(0\right)$v(0).
What is the discriminant of the equation $2\cos^2\left(t\right)-\cos t+4=0$2cos2(t)−cost+4=0?
Which two statements together justify the direction in which the particle moves?
The initial velocity is negative and the object never comes to rest.
The initial velocity is positive and the object never comes to rest.
The initial velocity is positive and the object comes to rest and turns around.
The initial velocity is negative and the object comes to rest and turns around.
Therefore, it is always moving in the positive direction.
Therefore, it is always moving in the negative direction.
The initial velocity is negative and the object never comes to rest.
The initial velocity is positive and the object never comes to rest.
The initial velocity is positive and the object comes to rest and turns around.
The initial velocity is negative and the object comes to rest and turns around.
Therefore, it is always moving in the positive direction.
Therefore, it is always moving in the negative direction.
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
Apply differentiation methods in solving problems