Motion and Rates (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Finding Displacement from Velocity

Let's consider a particle moving with velocity $v = 2t - 2$ m/s, where $t$ is measured in seconds. Initially (at $t = 0$), the particle is at position $x = 1$ metre.

Part (a): Find the displacement function

Start with the velocity function:

$v = 2t - 2 \quad \text{(equation 1)}$

Integrate to find displacement:

$x = t^2 - 2t + C$

where $C$ is the constant of integration.

Now use the initial condition. When $t = 0$, $x = 1$:

$1 = 0 - 0 + C$

Therefore $C = 1$.

The displacement function is:

$x = t^2 - 2t + 1$

which can be factorised as:

$x = (t - 1)^2 \quad \text{(equation 2)}$

Part (b): When is the particle at the origin?

The particle is at the origin when $x = 0$. Using equation 2:

$(t - 1)^2 = 0$

$t = 1$

The particle is at the origin when $t = 1$ second.

To find the velocity at this time, substitute $t = 1$ into equation 1:

$v = 2(1) - 2 = 0 \text{ m/s}$

The particle is stationary at the origin.

Part (c): Why is the particle never on the negative side of the origin?

Since $x = (t - 1)^2$, the value of $x$ is always a perfect square. Perfect squares are never negative, so $x \geq 0$ for all values of $t$. Therefore, the particle never has a negative displacement.

Part (d): Find the acceleration

To find acceleration, differentiate the velocity function $v = 2t - 2$:

$a = 2$

The acceleration is constant at $2$ m/s².

Worked Example: Finding Velocity and Displacement from Acceleration

Consider a particle with acceleration function $a = 24t$ cm/s². Initially, it is at the origin with velocity $-12$ cm/s.

Part (a): Find the velocity function

Start with the acceleration:

$a = 24t \quad \text{(equation 1)}$

Integrate to find velocity:

$v = 12t^2 + C$

Use the initial condition: when $t = 0$, $v = -12$:

$-12 = 0 + C$

Therefore $C = -12$.

The velocity function is:

$v = 12t^2 - 12 \quad \text{(equation 2)}$

Part (b): Find the displacement function

Integrate the velocity function:

$x = 4t^3 - 12t + D$

Use the initial condition: when $t = 0$, $x = 0$:

$0 = 0 - 0 + D$

Therefore $D = 0$.

The displacement function is:

$x = 4t^3 - 12t \quad \text{(equation 3)}$

Part (c): When is the particle stationary?

The particle is stationary when $v = 0$. Using equation 2:

$12t^2 - 12 = 0$

$t^2 = 1$

$t = 1$

(taking the positive value since $t \geq 0$)

The particle is stationary after $1$ second.

When $t = 1$:

$x = 4(1)^3 - 12(1) = -8$

At this time, the displacement is $-8$ cm.

Part (d): When does the particle return to the origin?

The particle returns to the origin when $x = 0$. Using equation 3:

$4t^3 - 12t = 0$

$4t(t^2 - 3) = 0$

So $t = 0$ or $t^2 = 3$

Taking the positive non-zero value: $t = \sqrt{3}$ seconds.

At this time, the acceleration is:

$a = 24\sqrt{3} \text{ cm/s}^2$

Worked Example: Stone Dropped from a Building

A stone is dropped from the top of a high building. How far has it travelled, and how fast is it going, after $5$ seconds? Take $g = 9.8$ m/s².

Setting up the problem

Let $x$ metres be the distance travelled $t$ seconds after the stone is dropped.

This choice means:

• The origin of space is at the top of the building
• The origin of time is when the stone is dropped
• Downwards is the positive direction
• Units are metres and seconds

Finding the velocity function

The acceleration is:

$a = 9.8 \quad \text{(equation 1)}$

Integrate to find velocity:

$v = 9.8t + C$

The stone was dropped (not thrown), so its initial speed was zero. When $t = 0$, $v = 0$:

$0 = 0 + C$

Therefore $C = 0$.

The velocity function is:

$v = 9.8t \quad \text{(equation 2)}$

Finding the displacement function

Integrate the velocity:

$x = 4.9t^2 + D$

The initial displacement was zero (we placed the origin at the top). When $t = 0$, $x = 0$:

$0 = 0 + D$

Therefore $D = 0$.

The displacement function is:

$x = 4.9t^2 \quad \text{(equation 3)}$

Calculating the answers

When $t = 5$:

From equation 2: $v = 9.8(5) = 49$ m/s

From equation 3: $x = 4.9(5)^2 = 122.5$ metres

Therefore, after $5$ seconds, the stone has fallen $122.5$ metres and is moving downwards at $49$ m/s.

Worked Example: Ball Thrown Upwards from a Lookout

A cricketer stands on a lookout that projects out over a valley floor $100$ metres below him. He throws a cricket ball vertically upwards at a speed of $40$ m/s, and it falls back past the lookout onto the valley floor below.

How long does it take to fall, and with what speed does it strike the ground? (Take $g = 10$ m/s².)

Setting up the problem

Let $x$ metres be the distance above the valley floor $t$ seconds after the ball is thrown.

This choice means:

• The origin of space is at the valley floor (not at the lookout)
• The origin of time is when the ball is thrown
• Upwards is positive
• Therefore, the acceleration is $a = -10$ (because gravity acts downwards)

Finding the velocity function

The acceleration is:

$a = -10 \quad \text{(equation 1)}$

Integrate to find velocity:

$v = -10t + C$

When $t = 0$, $v = 40$ (the initial upward velocity):

$40 = 0 + C$

Therefore $C = 40$.

The velocity function is:

$v = -10t + 40 \quad \text{(equation 2)}$

Finding the displacement function

Integrate the velocity:

$x = -5t^2 + 40t + D$

When $t = 0$, $x = 100$ (the cricketer is $100$ m above the valley floor):

$100 = 0 + 0 + D$

Therefore $D = 100$.

The displacement function is:

$x = -5t^2 + 40t + 100 \quad \text{(equation 3)}$

When does the ball hit the ground?

The ball hits the ground when $x = 0$. Using equation 3:

$-5t^2 + 40t + 100 = 0$

Divide by $-5$:

$t^2 - 8t - 20 = 0$

Factorise:

$(t - 10)(t + 2) = 0$

So $t = 10$ or $t = -2$

Since the ball was not in flight at $t = -2$, we take $t = 10$ seconds.

What is the impact speed?

Substitute $t = 10$ into equation 2:

$v = -10(10) + 40 = -60 \text{ m/s}$

The negative sign indicates downward motion (since we chose upwards as positive).

Therefore, the ball hits the ground after $10$ seconds at a speed of $60$ m/s downwards.

Worked Example: Trigonometric Velocity Function

The velocity of a particle initially at the origin is $v = \sin \frac{1}{4}t$, measured in metres and seconds.

Part (a): Find the displacement function

The velocity is:

$v = \sin \frac{1}{4}t \quad \text{(equation 1)}$

Integrate to find displacement:

$x = -4 \cos \frac{1}{4}t + C$

Use the initial condition: when $t = 0$, $x = 0$:

$0 = -4(1) + C$

Therefore $C = 4$.

The displacement function is:

$x = 4 - 4\cos \frac{1}{4}t \quad \text{(equation 2)}$

Part (b): Find the acceleration function

Differentiate the velocity function:

$a = \frac{1}{4}\cos \frac{1}{4}t \quad \text{(equation 3)}$

Part (c): Find displacement, velocity and acceleration when $t = 4\pi$

When $t = 4\pi$:

Using equation 2:

$x = 4 - 4\cos \pi = 4 - 4(-1) = 8 \text{ metres}$

Using equation 1:

$v = \sin \pi = 0 \text{ m/s}$

Using equation 3:

$a = \frac{1}{4}\cos \pi = -\frac{1}{4} \text{ m/s}^2$

Part (d): Describe the motion

The particle oscillates between $x = 0$ (when $\cos \frac{1}{4}t = 1$) and $x = 8$ (when $\cos \frac{1}{4}t = -1$).

The period of oscillation is $8\pi$ seconds (since the cosine function completes one full cycle when $\frac{1}{4}t$ increases by $2\pi$, which means $t$ increases by $8\pi$).

Worked Example: Exponential Acceleration Function

The acceleration of a particle is given by $a = e^{-2t}$ (in metres and seconds), and the particle is initially stationary at the origin.

Part (a): Find the velocity and displacement functions

The acceleration is:

$a = e^{-2t} \quad \text{(equation 1)}$

Integrate to find velocity:

$v = -\frac{1}{2}e^{-2t} + C$

When $t = 0$, $v = 0$ (initially stationary):

$0 = -\frac{1}{2} + C$

Therefore $C = \frac{1}{2}$.

The velocity function is:

$v = -\frac{1}{2}e^{-2t} + \frac{1}{2} \quad \text{(equation 2)}$

Integrate again to find displacement:

$x = \frac{1}{4}e^{-2t} + \frac{1}{2}t + D$

When $t = 0$, $x = 0$ (initially at origin):

$0 = \frac{1}{4} + D$

Therefore $D = -\frac{1}{4}$.

The displacement function is:

$x = \frac{1}{4}e^{-2t} + \frac{1}{2}t - \frac{1}{4} \quad \text{(equation 3)}$

Part (b): Find the displacement when $t = 10$

Using equation 3 when $t = 10$:

$x = \frac{1}{4}e^{-20} + 5 - \frac{1}{4}$

$x = 4\frac{3}{4} + \frac{1}{4}e^{-20} \text{ metres}$

(Note: $e^{-20}$ is an extremely small positive number, approximately $0.0000000021$)

Part (c): Describe the velocity behaviour

Using equation 2, we can analyse the velocity:

When $t = 0$: $v = 0$ (initially stationary)

As $t$ increases, $e^{-2t}$ approaches zero, so:

$v \to -\frac{1}{2}(0) + \frac{1}{2} = \frac{1}{2} \text{ m/s}$

The velocity starts at zero and increases, approaching a limiting velocity of $\frac{1}{2}$ m/s as time goes on.

This type of motion occurs when resistance forces (like air resistance or friction) increase with velocity, eventually balancing the driving force.