 iGCSE (2021 Edition)

# 11.21 Kinematics

Lesson

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 positions indicate a position to the right of the origin.

A displacement function, usually given by $x\left(t\right)$x(t) or $s\left(t\right)$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 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. This means its final position is $3.75$3.75 units to the left of the origin. A possible displacement function for the particle above is $x\left(t\right)=-0.75t^2+3t$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, after $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. Also note that distinct from speed, velocity also includes direction. When the graph above has a positive velocity (positive gradient) the particle is moving toward 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 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 $x\left(t\right)$x(t) then the velocity function, $v\left(t\right)$v(t), is given by:

$v\left(t\right)=x'\left(t\right)$v(t)=x(t)

For our example above the velocity function is $v\left(t\right)=-1.5t+3$v(t)=1.5t+3.

The rate of change of velocity is acceleration. So if an object has a velocity given by $v\left(t\right)$v(t), then the acceleration function $a\left(t\right)$a(t), is given by:

$a\left(t\right)=v'\left(t\right)$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. Hence, for a displacement function $x\left(t\right)$x(t), the acceleration function is given by:

$a\left(t\right)=x''\left(t\right)$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 $v\left(t\right)=-1.5t+3$v(t)=1.5t+3, so the acceleration function is $a\left(t\right)=-1.5$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 $x\left(t\right)=t^3-9t^2+24t-10$x(t)=t39t2+24t10 m from the origin, where $t$t is time in seconds, $t\ge0$t0.

(a) Find the initial position of the object.

Think: The initial position is when $t=0$t=0. Substitute this into the displacement function and interpret.

Do:

 $x\left(0\right)$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 for the object.

Think: The velocity function is the derivative of the displacement function.

Do:

 $v\left(t\right)$v(t) $=$= $x'\left(t\right)$x′(t) $=$= $3t^2-18t+24$3t2−18t+24

(c) At what times does the object change directions?

Think: The object changes directions when there is a turning point in $x\left(t\right)$x(t) or when the velocity is zero (you have to stop to turn around!). So we need to solve $v\left(t\right)=0$v(t)=0 and show that the velocity changes sign or test concavity (that is we have a local maximum or minimum and not a stationary point of inflection).

Do:

 $3t^2-18t+24$3t2−18t+24 $=$= $0$0 $3\left(t^2-6t+8\right)$3(t2−6t+8) $=$= $0$0 $3\left(t-2\right)\left(t-4\right)$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
$v\left(t\right)$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 for the object.

Think: The acceleration function is the derivative of the velocity function.

Do:

 $a\left(t\right)$a(t) $=$= $v'\left(t\right)$v′(t) $=$= $6t-18$6t−18

(e) Describe the motion of the object at $t=1$t=1 second.

Think: To describe the motion, find the object's position, velocity and acceleration and interpret.

Do:

 Position Velocity Acceleration $x\left(1\right)$x(1) $=$= $\left(1\right)^3-9\left(1\right)^2+24\left(1\right)-10$(1)3−9(1)2+24(1)−10 $v\left(1\right)$v(1) $=$= $3\left(1\right)^2-18+24$3(1)2−18+24 $a\left(1\right)$a(1) $=$= $6\left(1\right)-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.

Summary

The displacement function, $x\left(t\right)$x(t), gives the position of a particle from the origin at time $t$t.

The velocity function, $v\left(t\right)$v(t), gives the instantaneous rate of change of the displacement function with respect to time.

$v\left(t\right)=x'\left(t\right)$v(t)=x(t)

The acceleration function, $a\left(t\right)$a(t), gives the instantaneous rate of change of the velocity with respect to time.

 $a\left(t\right)$a(t) $=$= $v'\left(t\right)$v′(t) $=$= $x''\left(t\right)$x′′(t)

Interpretation:

• $x\left(t\right)>0$x(t)>0, the particle is located to the right of the origin.
• $x\left(t\right)<0$x(t)<0, the particle is located to the left of the origin.
• $x\left(t\right)=0$x(t)=0, the particle is at the origin.

• $v\left(t\right)>0$v(t)>0, the particle is moving to the right
• $v\left(t\right)<0$v(t)<0, the particle is moving to the left
• $v\left(t\right)=0$v(t)=0, the particle is instantaneously at rest. If the sign of $v\left(t\right)$v(t) changes at this point, then the particle changes direction at this point.

• $a\left(t\right)>0$a(t)>0, the particle is accelerating to the right
• $a\left(t\right)<0$a(t)<0, the particle is accelerating to the left
• $a\left(t\right)=0$a(t)=0, the particle has a constant instantaneous velocity at this point

• If $v\left(t\right)$v(t) and $a\left(t\right)$a(t) have the same signs, the particle is increasing in speed
• If $v\left(t\right)$v(t) and $a\left(t\right)$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 $x\left(t\right)=18\sqrt{t}$x(t)=18t.

1. 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.

2. Hence, calculate the velocity of the object after $9$9 seconds.

##### Question 2

A particle moves in a straight line and its displacement after $t$t seconds is given by $x=12t-2t^2$x=12t2t2, 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.

1. Determine an equation for the velocity $v$v of the particle after $t$t seconds.

2. After how many seconds $t$t does the particle change its direction of motion?

Write each line of working as an equation.

3. Plot the graph of displacement v.s. time on the axes below:

4. Determine the displacement of the particle after $9$9 seconds.

5. Hence find the total distance that the particle has traveled in the first $9$9 seconds.

##### Question 3

A particle moves in a straight line, starting from rest at the point $O$O. Its displacement is positive when moving to the right, and negative when moving to the left.

At time $t$t seconds after leaving $O$O, the speed $v$v of the particle (in m/s) is given by $v\left(t\right)=\left(3t-2t^2\right)^2$v(t)=(3t2t2)2.

1. Solve for the time $t>0$t>0 at which the particle instantaneously comes to rest.

Write each line of working as an equation.

2. Determine $a\left(2\right)$a(2), the acceleration of the particle after $2$2 seconds.

Write each line of working as an equation.

3. Which of the following correctly describes the motion of the particle?

The particle starts moving to the right, but eventually comes to rest. After instantaneously coming to rest, it turns around and starts moving to the left.

A

The particle starts moving to the right, but eventually comes to rest. After instantaneously coming to rest, it then continues to move to the right.

B

The particle starts moving to the right, never slowing down as it continues to move further and further to the right.

C

The particle starts moving to the right, but eventually comes to rest. After instantaneously coming to rest, it turns around and starts moving to the left.

A

The particle starts moving to the right, but eventually comes to rest. After instantaneously coming to rest, it then continues to move to the right.

B

The particle starts moving to the right, never slowing down as it continues to move further and further to the right.

C

##### Question 4

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.

The displacement of an object $x$x about a fixed position $x=0$x=0, at time $t$t seconds, is given by $x\left(t\right)=-5\cos2t$x(t)=5cos2t. Its velocity at time $t$t seconds is given by $v\left(t\right)$v(t) and its acceleration at time $t$t seconds is given by $a\left(t\right)$a(t).

1. Show that $a\left(t\right)=-4x\left(t\right)$a(t)=4x(t).

2. Determine the acceleration of the object when it is at the origin $x=0$x=0.

##### Question 5

The displacement (in metres) of an object along a straight line after $t$t seconds is modelled by $x\left(t\right)=4te^{-2t}$x(t)=4te2t.

1. Find the velocity function for the object, $v\left(t\right)$v(t).

2. Find the acceleration function for the object, $a\left(t\right)$a(t).

3. Find the object's initial displacement in metres.

4. At what value of $t$t does the object change direction?

5. Find the object's initial acceleration in m/s2.

6. At what value of $t$t is the object travelling at the greatest speed towards the origin?

### Outcomes

#### 0606C14.11

Apply differentiation and integration to kinematics problems that involve displacement, velocity and acceleration of a particle moving in a straight line with variable or constant acceleration, and the use of x–t and v–t graphs.