What is the difference between displacement and distance?

Displacement is measured relative to a fixed point and may be positive, negative or zero. Distance is never negative, and often means the length actually travelled. A bus back at its depot has zero displacement, but the distance it has travelled is the whole route.

A particle is the general term for the moving object in a kinematics problem. It is treated as being the size of a single point, so its shape and dimensions can be ignored entirely.

Which quantity tells you which way the particle is moving?

The sign of the velocity , in every case. A positive velocity means motion in the positive direction and a negative one means motion in the negative direction; the acceleration's sign does not answer this question.

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

False. You have to compare it with the velocity: matching signs mean the particle is speeding up, and opposite signs mean it is slowing down. So a negative acceleration with a negative velocity is a particle speeding up in the negative direction.

What do velocity and acceleration mean on a graph?

Velocity is the gradient of a displacement-time graph. Acceleration is the gradient of a velocity-time graph, which is the same relationship applied one step further along.

True or False? Getting from acceleration to displacement needs two constants of integration.

True. Integrating the acceleration gives the velocity and one constant, and integrating again gives the displacement and a second. Each needs its own piece of information, so a question of this kind must supply two known values.

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Calculus for Kinematics (Edexcel IGCSE Further Pure Maths): Flashcards

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What is the difference between displacement and distance?

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What is the difference between displacement and distance?
Displacement is measured relative to a fixed point and may be positive, negative or zero.
Distance is never negative, and often means the length actually travelled.
A bus back at its depot has zero displacement, but the distance it has travelled is the whole route.
What is the difference between speed and velocity?
Velocity carries a sign, which tells you the direction of travel.
Speed is its magnitude, $|v|$ , so a velocity of $- 6$ is a speed of $6$ .
Define particle.
A particle is the general term for the moving object in a kinematics problem.
It is treated as being the size of a single point, so its shape and dimensions can be ignored entirely.
What does it mean for a particle to be at rest?
Its velocity is zero.
Stationary and instantaneously at rest both say the same thing, that $v = 0$ at that moment.
Which quantity tells you which way the particle is moving?
The sign of the velocity, in every case.
A positive velocity means motion in the positive direction and a negative one means motion in the negative direction; the acceleration's sign does not answer this question.
What does $a = 0$ tell you about the motion?
The particle is moving with constant velocity.
That does not mean it has stopped: it simply is not speeding up or slowing down at that moment.
True or False?
A negative acceleration means the particle is slowing down.
False.
You have to compare it with the velocity: matching signs mean the particle is speeding up, and opposite signs mean it is slowing down.
So a negative acceleration with a negative velocity is a particle speeding up in the negative direction.
Complete the two derivatives linking displacement, velocity and acceleration:
$v = \frac{d______}{d t} and a = \frac{d______}{d t} = \frac{d^{2} s}{d t^{2}}$
The completed derivatives are:
$v = \frac{d s}{d t} and a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$
Differentiating moves you down the chain from displacement to velocity to acceleration.
What do velocity and acceleration mean on a graph?
Velocity is the gradient of a displacement-time graph.
Acceleration is the gradient of a velocity-time graph, which is the same relationship applied one step further along.
Given $s (t) = 2 t^{3} - 18 t^{2} + 48 t$ , find $v$ and $a$ .
Differentiating once gives $v (t) = 6 t^{2} - 36 t + 48$ .
Differentiating again gives $a (t) = 12 t - 36$ .
Powers of $t$ behave exactly as powers of $x$ do, so no new technique is involved.
How do you find when a particle is instantaneously at rest?
Solve $v (t) = 0$ .
For $v (t) = 6 t^{2} - 36 t + 48$ that factorises to $6 (t - 2) (t - 4) = 0$ , so the particle is at rest at $t = 2$ and $t = 4$ .
How do you find the minimum velocity, and why is it not the minimum speed?
Solve $\frac{d v}{d t} = 0$ , which is $a = 0$ , then substitute that time back into $v$ .
For the motion above $12 t - 36 = 0$ gives $t = 3$ and $v (3) = - 6 m s^{- 1}$ .
The minimum speed is $0$ , reached at $t = 2$ and $t = 4$ , because speed ignores the sign.
Complete the two integrals that reverse the chain:
$s = \int______d t and v = \int______d t$
The completed integrals are:
$s = \int v d t and v = \int a d t$
Integrating moves you back up the chain, in the opposite direction to differentiating.
What extra information finds the constant in a kinematics problem?
The value of $s$ or $v$ at one stated time.
Watch for the word initially, which always means $t = 0$ , and substitute that value into your integrated expression to solve for $c$ .
A particle has $v (t) = 6 t^{2} - t + 3$ and $s = - 5$ when $t = 0$ ; find $s$ when $t = 4$ .
Integrating gives $s (t) = 2 t^{3} - \frac{1}{2} t^{2} + 3 t + c$ , and substituting $t = 0$ shows $c = - 5$ .
Then $s (4) = 128 - 8 + 12 - 5 = 127 m$ .
True or False?
Getting from acceleration to displacement needs two constants of integration.
True.
Integrating the acceleration gives the velocity and one constant, and integrating again gives the displacement and a second.
Each needs its own piece of information, so a question of this kind must supply two known values.

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