Kinematics is the study of the motion of objects. Here we will consider the movement of a particle in a straight line. The term particle refers to a point which in context may be the centre of the object of interest. When describing the motion of a particle we will make use of the concepts of displacement, velocity and acceleration.
Position of an object in straight line motion is its location relative to an origin. Commonly we will consider the point moving along a horizontal line with given position $x$x, where $x=0$x=0 is the origin, negative values of $x$x indicate a position to the left of the origin and positive positions indicate a position to the right of the origin.
A displacement function, usually given by $x(t)$x(t) or $s(t)$s(t), describes the particle's change in position and will give the particle's position at time $t$t. In contrast to distance, displacement is a vector quantity and has direction.
For example, if a particle has initial position $x_1=0$x1=0, that is it starts at the origin and the moves $3$3 units to the right (position $x_2$x2) followed by $6.75$6.75 units to the left to be in a final position $x_3$x3. Illustrated in the motion diagram below the particle has travelled a total distance of $9.75$9.75 units but has displacement $-3.75$−3.75 units. That is its final position is $3.75$3.75 units to the left of the origin.
A possible displacement function for the particle above is $x(t)=-0.75t^2+3t$x(t)=−0.75t2+3t, where $x$x is the position in metres after $t$t seconds. This gives us the following position-time graph:
We can see the particle started at the origin, then moved to the right, at $2$2 seconds it reached position $x_2$x2 at $3$3 m to the right of the origin. The particle then started moving to the left, passing the origin at $4$4 seconds and coming to the final position $x_3=-3.75$x3=−3.75 m at $t=5$t=5 seconds. At the turning point the particle changed direction.
What does the gradient of a tangent to a position-time graph represent? The rate of change of displacement over time is the velocity of the particle. For the example above, we have units of metres per second, distinct from speed velocity also has direction. When the graph above has a positive velocity(positive gradient) the particle is moving towards the right and when the graph have a negative velocity(negative gradient) the particle is moving toward the left.
Turning points (local maxima or minima) indicate a change in direction and an instantaneous rate of change of $0$0.
As the velocity is the rate of change of the displacement we can use differentiation to calculate the instantaneous velocity of a particle at any time $t$t. If we represent the displacement function with $x(t)$x(t) then the velocity function, $v(t)$v(t), is given by:
$v(t)=x'(t)$v(t)=x′(t)
For our example above the velocity function is $v(t)=-1.5t+3$v(t)=−1.5t+3.
The rate of change of velocity is acceleration. So if an object has a velocity given by $v(t)$v(t), then the acceleration function $a(t)$a(t), is given by:
$a(t)=v'(t)$a(t)=v′(t)
We could also describe this in terms of the displacement function. To obtain the acceleration function from the displacement function we would need to differentiate twice. This is called finding the second derivative, for a function $f(x)$f(x) its second derivative is denoted $f''(x)$f′′(x). You may be able to guess what the notation for the third and fourth derivative is. For Leibniz's notation the second derivative of $y=f(x)$y=f(x) is $\frac{d^2y}{dx^2}$d2ydx2 which reads "take the derivative twice of the function $y$y, with respect to $x$x both times".
Hence, for a displacement function $x(t)$x(t), the acceleration function is given by:
$a(t)=x''(t)$a(t)=x′′(t)
As acceleration is the rate of change of velocity the units of acceleration are the units for velocity over time, which is equivalent to the unit for displacement divided by the unit of time squared.
For our previous example we had $v(t)=-1.5t+3$v(t)=−1.5t+3, so the acceleration function is $a(t)=-1.5$a(t)=−1.5 m/s2. This indicates a constant acceleration of $1.5$1.5 m/s2 to the left.
Worked example
The displacement of an object moving in a straight line is given by $x(t)=t^3-9t^2+24t-10$x(t)=t3−9t2+24t−10 m from the origin, where $t$t is time in seconds, $t\ge0$t≥0.
a) Find the initial position of the object.
b) Find the velocity function for the object.
c) At what times does the object change directions?
d) Find the acceleration function for the object.
e) Describe the motion of the object at $t=1$t=1 second.
Solution:
a) The initial position is when $t=0$t=0:
| $x\left(0\right)$x(0) | $=$= | $\left(0\right)^3-9\left(0\right)^2+24\times\left(0\right)-10$(0)3−9(0)2+24×(0)−10 |
| $=$= | $-10$−10 |
Thus, the object is initially located $10$10 m to the left of the origin.
b) Find the velocity function using differentiation.
| $v(t)$v(t) | $=$= | $x'(t)$x′(t) |
| $=$= | $3t^2-18t+24$3t2−18t+24 |
c) The object changes directions when there is a turning point in $x(t)$x(t). So we need to solve $v(t)=0$v(t)=0 and show that the velocity changes sign (that is we have a local maximum or minimum and not a stationary point of inflection).
| $3t^2-18t+24$3t2−18t+24 | $=$= | $0$0 |
| $3(t^2-6t+8)$3(t2−6t+8) | $=$= | $0$0 |
| $3(t-2)(t-4)$3(t−2)(t−4) | $=$= | $0$0 |
| $\therefore t$∴t | $=$= | $2$2 or $4$4 |
Hence, there are stationary points at $t=2$t=2 and $t=4$t=4. Show the nature of the stationary points to confirm a change in direction.
| $t$t | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|
| $v(t)$v(t) | $9$9 | $0$0 | $-3$−3 | $0$0 | $9$9 |
| Sign | $+$+ | $0$0 | $-$− | $0$0 | $+$+ |
| Shape |  |
 |
 |
|
|
We have confirmed that the velocity changed signs either side of the stationary points, hence, the object changed directions at $t=2$t=2 seconds and $t=4$t=4 seconds.
d) Find the acceleration function using differentiation on the velocity function.
| $a(t)$a(t) | $=$= | $v'(t)$v′(t) |
| $=$= | $6t-18$6t−18 |
e) To describe the motion, find the object's position, velocity and acceleration and interpret.
| Position | Velocity | Acceleration | ||||||||
| $x(1)$x(1) | $=$= | $(1)^3-9(1)^2+24(1)-10$(1)3−9(1)2+24(1)−10 | $v(1)$v(1) | $=$= |
$3(1)^2-18+24$3(1)2−18+24 |
$a(1)$a(1) | $=$= | $6(1)-18$6(1)−18 | ||
| $=$= | $6$6 m | $=$= | $9$9 m/s | $=$= | $-12$−12 m/s2 |
Interpretation: At $t=1$t=1 second the object is located at $6$6 m to the right of the origin and is travelling at a velocity of $9$9 m/s to the right and accelerating at $12$12 m/s2 to the left. As the acceleration and velocity are in different directions (different signs) the object is slowing down.
The displacement function, $x(t)$x(t), gives the position of a particle from the origin at time $t$t.
The velocity function, $v(t)$v(t), gives the instantaneous rate of change of the displacement function with respect to time.
$v(t)=x'(t)$v(t)=x′(t)
The acceleration function, $a(t)$a(t), gives the instantaneous rate of change of the velocity with respect to time.
$a(t)=v'(t)$a(t)=v′(t)
Interpretation:
- $x(t)>0$x(t)>0, the particle is located to the right of the origin.
- $x(t)<0$x(t)<0, the particle is located to the left of the origin.
- $x(t)=0$x(t)=0, the particle is at the origin.
- $v(t)>0$v(t)>0, the particle is moving to the right
- $v(t)<0$v(t)<0, the particle is moving to the left
- $v'(t)=0$v′(t)=0, the particle is instantaneously at rest. If the sign of $v(t)$v(t) changes at this point, then the particle changes direction at this point.
- $a(t)>0$a(t)>0, the particle is accelerating to the right
- $a(t)<0$a(t)<0, the particle is accelerating to the left
- $a(t)=0$a(t)=0, the particle has a constant instantaneous velocity at this point
- If $v(t)$v(t) and $a(t)$a(t) have the same signs, the particle is increasing in speed
- If $v(t)$v(t) and $a(t)$a(t) have opposite signs, the particle is decreasing in speed
Practice questions
Question 1
The position (in metres) of an object along a straight line after $t$t seconds is modelled by $x\left(t\right)=6t^2$x(t)=6t2.
State the velocity $v\left(t\right)$v(t) of the particle at time $t$t.
Which of the following represent the velocity of the particle after $4$4 seconds? Select all that apply.
$x'\left(4\right)$x′(4)
A$v'\left(4\right)$v′(4)
B$x\left(4\right)$x(4)
C$v\left(4\right)$v(4)
DHence find the velocity of the particle after $4$4 seconds.
Question 2
The position (in metres) of an object along a straight line after $t$t seconds is modelled by $x\left(t\right)=18\sqrt{t}$x(t)=18√t.
Determine the function $v\left(t\right)$v(t) for the velocity of the particle. Express $v\left(t\right)$v(t) in surd form.
Hence, calculate the velocity of the object after $9$9 seconds.
Question 3
The displacement of a particle moving in rectilinear motion is given by $x\left(t\right)=-5t\left(t-4\right)$x(t)=−5t(t−4) where $x$x is the displacement in metres from the origin and $t$t is the time in seconds.
Calculate the initial displacement of the particle.
Solve for the time $t$t when the particle next returns to the origin.
Using graphical methods, calculate the distance traveled by the particle between leaving the origin and returning again.
Question 4
The velocity, in m/s, of a body moving in rectilinear motion is given by $v\left(t\right)=5t^2-23t+24$v(t)=5t2−23t+24 where $t$t is time in seconds.
Solve for the time(s), $t$t, when the body is instantaneously at rest.
If the acceleration of a particle is the rate of change of the velocity, determine the function $a\left(t\right)$a(t) for the acceleration of the body.
Calculate the acceleration at time $t=3$t=3.