Define displacement .

Displacement is the distance of a particle from a fixed point, measured relative to that point. It can be negative, meaning the particle is on the negative side of the fixed point, and it is zero whenever the particle returns there.

A bus leaves its depot, drives a full route and returns. What are its displacement and its distance travelled?

Its displacement is zero , because it has ended where it began. The distance travelled is the whole length of the route, because distance ignores direction and is always positive.

True or False? A particle with negative acceleration is slowing down.

False. What matters is whether velocity and acceleration have the same sign : if they do the particle speeds up, and if they differ it slows down. A particle with negative velocity and negative acceleration is speeding up, in the negative direction.

What does a horizontal line on a velocity-time graph mean?

The velocity is constant , so the acceleration is zero. Above the axis the particle moves forwards at that steady velocity, and below it moves backwards.

On a velocity-time graph, how do you get the distance rather than the displacement?

Add all the areas, treating those below the axis as positive. Displacement instead subtracts the areas below the axis from those above, which is why the two answers differ whenever the particle turns round.

A particle is projected vertically upwards. What is its velocity at the highest point?

It is zero , since the particle is momentarily at rest as it changes direction. On a velocity-time graph this is where the line crosses the horizontal axis.

What does the gradient of a displacement-time graph give?

The velocity , since velocity is the rate of change of displacement. Its units are metres per second, which is exactly what a gradient of metres against seconds produces.

A displacement-time graph is a cubic. What shapes are the velocity-time and acceleration-time graphs?

The velocity-time graph is a quadratic and the acceleration-time graph is linear . Each differentiation drops the degree by one, so the three sketches follow from the first.

A velocity-time graph is a straight line with negative gradient. What does the acceleration-time graph look like?

A horizontal line below the time axis . A straight velocity graph has constant gradient, so the acceleration is constant, and a negative gradient puts that constant below the axis.

True or False? The area under a displacement-time graph gives a useful quantity.

False. It has no meaningful interpretation, unlike the areas under velocity-time and acceleration-time graphs. On a displacement-time graph it is the gradient that carries the information.

IGCSEFurther MathsCambridge (CIE)Additional MathsFlashcardsCalculusCalculus for Kinematics

Calculus for Kinematics (Cambridge (CIE) IGCSE Additional Maths): Flashcards

Exam code: 0606

1/23

0Still learning

Know0

FrontKinematics Toolkit
Define displacement.

Was this flashcard helpful?

Cards in this collection (23)

Define displacement.
Displacement is the distance of a particle from a fixed point, measured relative to that point.
It can be negative, meaning the particle is on the negative side of the fixed point, and it is zero whenever the particle returns there.
A bus leaves its depot, drives a full route and returns. What are its displacement and its distance travelled?
Its displacement is zero, because it has ended where it began.
The distance travelled is the whole length of the route, because distance ignores direction and is always positive.
What is the difference between velocity and speed?
Velocity is the rate of change of displacement and carries a sign showing the direction of travel.
Speed is its magnitude, $|v|$ , so a velocity of $- 6 {m s}^{- 1}$ is a speed of $6 {m s}^{- 1}$ .
What does it mean for a particle to be "at rest"?
Its velocity is zero, so $v = 0$ .
That is not the same as being at the origin, which would be $s = 0$ .
True or False?
A particle with negative acceleration is slowing down.
False.
What matters is whether velocity and acceleration have the same sign: if they do the particle speeds up, and if they differ it slows down.
A particle with negative velocity and negative acceleration is speeding up, in the negative direction.
What does a horizontal line on a velocity-time graph mean?
The velocity is constant, so the acceleration is zero.
Above the axis the particle moves forwards at that steady velocity, and below it moves backwards.
On a velocity-time graph, how do you get the distance rather than the displacement?
Add all the areas, treating those below the axis as positive.
Displacement instead subtracts the areas below the axis from those above, which is why the two answers differ whenever the particle turns round.
A particle is projected vertically upwards. What is its velocity at the highest point?
It is zero, since the particle is momentarily at rest as it changes direction.
On a velocity-time graph this is where the line crosses the horizontal axis.
Complete the two derivative relationships:
$v =______$
$a =______$
The completed relationships are:
$v = \frac{d s}{d t}$
$a = \frac{d v}{d t}$
Velocity is the rate of change of displacement, and acceleration the rate of change of velocity.
How is acceleration related to displacement directly?
It is the second derivative, $a = \frac{d^{2} s}{d t^{2}}$ .
Differentiating displacement once gives velocity, and differentiating again gives acceleration.
A particle has displacement $x (t) = 2 t^{3} - 27 t^{2} + 84 t$ . Find its velocity.
Differentiate with respect to $t$ :
$x' (t) = 6 t^{2} - 54 t + 84 = 6 (t - 2) (t - 7)$
Factorised form is useful here, since it shows the particle is instantaneously at rest at $t = 2$ and $t = 7$ .
What does $s'' (t)$ represent, and what does $s' (t) = 0$ mean?
$s'' (t)$ is the acceleration, since displacement has been differentiated twice.
$s' (t) = 0$ means the velocity is zero, so the particle is momentarily at rest.
Complete the two integral relationships:
$s =______$
$v =______$
The completed relationships are:
$s = \int v d t$
$v = \int a d t$
Integration runs the chain backwards, from acceleration to velocity to displacement.
How do you find the constant of integration in a kinematics problem?
Use an initial or boundary condition given in the question.
The words "initially" or "initial" mean $t = 0$ , so substituting those values produces an equation for the constant.
What does $\int_{t_{1}}^{t_{2}} v (t) d t$ give you?
The displacement of the particle between those two times.
Any interval where the velocity is negative contributes a negative amount, so this is the net change in position.
True or False?
$\int_{t_{1}}^{t_{2}} v (t) d t$ gives the distance travelled.
False.
That integral gives the displacement; for the distance you need $\int_{t_{1}}^{t_{2}} |v (t)| d t$ .
The two agree only while the particle keeps moving one way, and differ as soon as it turns round.
Fill in the two operations that move between the three quantities:
Going from $s$ to $v$ to $a$ you $______$ , and going from $a$ to $v$ to $s$ you $______$ .
The completed sentence is:
Going from $s$ to $v$ to $a$ you differentiate, and going from $a$ to $v$ to $s$ you integrate.
Each integration brings in a constant, so going backwards needs extra information each time.
On a velocity-time graph, what do the gradient and the area represent?
The gradient is the acceleration and the area is the displacement.
The units confirm it: a gradient in ${m s}^{- 1}$ per second is ${m s}^{- 2}$ , and an area in ${m s}^{- 1} \times s$ is metres.
What does the gradient of a displacement-time graph give?
The velocity, since velocity is the rate of change of displacement.
Its units are metres per second, which is exactly what a gradient of metres against seconds produces.
What does the area under an acceleration-time graph give?
The velocity, because integrating acceleration with respect to time gives velocity.
Again the units agree: ${m s}^{- 2} \times s$ gives ${m s}^{- 1}$ .
A displacement-time graph is a cubic. What shapes are the velocity-time and acceleration-time graphs?
The velocity-time graph is a quadratic and the acceleration-time graph is linear.
Each differentiation drops the degree by one, so the three sketches follow from the first.
A velocity-time graph is a straight line with negative gradient. What does the acceleration-time graph look like?
A horizontal line below the time axis.
A straight velocity graph has constant gradient, so the acceleration is constant, and a negative gradient puts that constant below the axis.
True or False?
The area under a displacement-time graph gives a useful quantity.
False.
It has no meaningful interpretation, unlike the areas under velocity-time and acceleration-time graphs.
On a displacement-time graph it is the gradient that carries the information.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Integration