26.06 Kinematics (Extended)
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.
The 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 values of $x$x indicate a position to the right of the origin.
Displacement
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, consider a particle that has initial position $x_1=0$x1=0 and then 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. The turning point indicates when the particle changed direction.
Velocity
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, velocity has units of metres per second and it also has direction. When the graph above has a positive velocity (positive gradient) the particle is moving towards the right and when the graph has 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 (velocity) 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.
Acceleration
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)
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 or m/s2.
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
example 1
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 |  |
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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.
Question 5
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:
Loading Graph...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.
Velocity graphs
Velocity graphs graph the velocity, $v$v, against time, $t$t.
- If the velocity function is increasing from left to right, then the particle is accelerating.
-
If the velocity function is decreasing from left to right, then the particle is decelerating.
-
If the velocity function is horizontal, then the particle is neither accelerating or decelerating.
Worked example
example 2
Consider the given velocity function for a particle and determine:
a) In what time intervals is the particle accelerating.
b) In what time intervals is the particle decelerating.
a) Think: For which values of t does the function have a positive gradient.
Do: $0\le t\le2,7\le t\le9$0≤t≤2,7≤t≤9
b) Think: For which values of t does the function have a negative gradient.
Do: $5\le t\le7$5≤t≤7
Displacement from Velocity
To find the net displacement from a velocity function, we can find the area between the velocity function and the $x$x-axis over a given interval. Net displacement may include both positive and negative values, as the velocity function accounts for both forward and backward movement. If the area is under the x-axis, than that means the displacement over that interval is negative. Total distance travelled, on the other hand, is always positive. To find the total distance travelled by an object, regardless of direction, we need to add all the areas between the x-axis and the velocity function.
Given the velocity function, $v\left(t\right)$v(t), if we wish to calculate the net change in displacement between $t=a$t=a and $t=b$t=b we have:
Net Displacement = the sum of the areas between the velocity function and the horizontal axis that are above the horizontal axis minus the sum of the areas between the velocity function and the horizontal axis that are below the horizontal axis
To calculate the distance travelled, $t=a$t=aand $t=b$t=b we have:
Distance travelled = the sum of the areas between the velocity function and the horizontal axis
Worked example
Example 3
A particle moves such that its velocity is given by $v(t)=3t-4$v(t)=3t−4. Find:
a) The displacement of the particle during the first four seconds.
b) The total distance travelled of the particle during the first four seconds.
a)
Think: We need to graph the velocity function so that we can find the areas between the function and the horizontal axis.
Do: Graph the linear function $v=3t-4$v=3t−4 using any appropriate method. e.g. Start at the vertical intercept: $c=-4$c=−4, and use the gradient: $m=\frac{rise}{run}=\frac{3}{1}$m=riserun=31 to find another point: $(1,-1)$(1,−1)then draw the line passing through these points:
Now we can find the areas between the function and the horizontal axis. They are shaded in the diagram below:
To find the area of each triangle, we will need to find the value of the $t$t-intercept in order to find the base length of each triangle. To do this, we substitute $v=0$v=0 into the equation:
| $v$v | $=$= | $3t-4$3t−4 |
| $0$0 | $=$= | $3t-4$3t−4 |
| $4$4 | $=$= | $3t$3t |
| $t$t | $=$= | $\frac{4}{3}$43 |
Now we can find the area of each triangle:
| Area of left triangle | $=$= |
$\frac{1}{2}\times\frac{4}{3}\times4$12×43×4 |
| $=$= | $\frac{8}{3}$83 | |
| Area of right triangle | $=$= | $\frac{1}{2}\times\left(4-\frac{4}{3}\right)\times8$12×(4−43)×8 |
| $=$= | $\frac{32}{3}$323 |
Now we can find the displacement by subtracting the area under the horizontal axis from the area above the horizontal axis:
| Displacement | $=$= | $\frac{32}{3}-\frac{8}{3}$323−83 |
| $=$= |
$8$8 units |
b) Now we can find the distance travelled by adding the areas under the horizontal axis and the area above the horizontal axis:
| Distance | $=$= | $\frac{32}{3}+\frac{8}{3}$323+83 |
| $=$= |
$\frac{40}{3}$403 units |