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The first and second derivatives of a position function read as velocity and acceleration, the speed as their length, and the two integrations with two initial conditions that recover a motion from the forces on it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to differentiate a position function to get velocity, speed and acceleration, integrate a known acceleration back to a motion using initial conditions in the right order, and split a projectile problem into its independent horizontal and vertical parts. You will also be able to say why a particle moving at constant speed can still be accelerating.
The derivative of a vector function from the last lesson, taken one component at a time, and the fact that a vector of constant length is perpendicular to its own derivative. This lesson gives those derivatives their physical names and asks the question in reverse: given the acceleration, what was the motion?
Position is $\mathbf{r}(t)$, velocity is $\mathbf{v} = \mathbf{r}'$, speed is the number $|\mathbf{v}|$, and acceleration is $\mathbf{a} = \mathbf{v}' = \mathbf{r}''$. Velocity and acceleration are vectors; speed is a scalar, and the difference between the second and the third of these is where most of the trouble in this lesson lives. A projectile is a particle whose only acceleration is a constant one, usually downward.
Let $\mathbf{r}(t)$ be the position of a particle at time $t$. Then
$$\mathbf{v}(t) = \mathbf{r}'(t), \qquad \text{speed} = |\mathbf{v}(t)|, \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t).$$
All three are computed one component at a time, so a motion in space is three motions on a line, carried together.
Velocity is not speed. The velocity is an arrow: it says where the particle is going and how fast. The speed is its length: one number, no direction. Two particles can have the same speed and opposite velocities, and a particle whose speed never changes can still be accelerating hard — which is what the last activity of this lesson is about.
The road back. Differentiation loses information, and each integration hands back one unknown constant vector:
$$\mathbf{v}(t) = \int \mathbf{a}(t)\,dt + \mathbf{C}_1, \qquad \mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C}_2.$$
$\mathbf{C}_1$ is fixed by the initial velocity and $\mathbf{C}_2$ by the initial position. Use each condition as soon as its constant appears: integrating twice before using either leaves two unknown vectors tangled in one expression.
Constant acceleration. When $\mathbf{a}$ does not depend on $t$ the two integrals are immediate:
$$\mathbf{v}(t) = \mathbf{v}_0 + t\,\mathbf{a}, \qquad \mathbf{r}(t) = \mathbf{r}_0 + t\,\mathbf{v}_0 + \tfrac{1}{2}t^{2}\,\mathbf{a}.$$
These are the school formulas for a projectile, written once for all three axes at the same time. With $\mathbf{a}$ pointing straight down, the horizontal components of $\mathbf{v}$ never change at all, and the whole of projectile motion is that one observation.
Another way: picture
Draw the curve and, at one point of it, two arrows from the same place. The velocity lies along the curve, pointing forward. The acceleration generally does not: it leans towards the inside of any bend, and only lines up with the velocity when the path is straight. Watching those two arrows as the particle moves tells you everything — velocity along the road, acceleration leaning towards wherever the road is turning.
Another way: steps
To go from a motion to its velocity and acceleration:
And to go the other way:
Take $\mathbf{a} = \langle 0, 0, -g \rangle$, a constant downward pull, with $\mathbf{r}_0 = \mathbf{0}$ and $\mathbf{v}_0 = \langle v_1, 0, v_3 \rangle$. Then
$$\mathbf{r}(t) = \left\langle v_1 t,\; 0,\; v_3 t - \tfrac{1}{2}g t^{2} \right\rangle.$$
The first component grows steadily and the third is a parabola in $t$, and neither depends on the other. The horizontal motion does not know it is falling; the vertical motion does not know it is moving along.
So every projectile question splits. When does it land? is a vertical question: set the third component to zero. How far does it go? is then a horizontal question asked at that time. When is it highest? is vertical again: set the third component of the velocity to zero, which happens at $t = v_3/g$.
At that highest moment the particle is not at rest — its speed is $|v_1|$, moving purely sideways. A learner who says the stone stops at the top has confused the vertical component of velocity with the velocity itself, and that is the same confusion as reading acceleration off a speedometer.
Differentiating the speed to get the acceleration. $|\mathbf{v}|'$ is a number and $\mathbf{a}$ is a vector, and they are not two spellings of one quantity. $|\mathbf{v}|'$ is only the part of $\mathbf{a}$ along the direction of travel; lesson 12 names the rest of it.
Substituting the time before differentiating. Putting $t = 3$ into $\mathbf{r}$ and then differentiating gives the zero vector every time.
Using both initial conditions at the end. Integrate, fix, integrate, fix. Two unfixed constant vectors cannot be separated by one condition applied afterwards.
Assuming acceleration points along the motion. It does so only on a straight path. On any bend it leans inward, and on a curve traced at constant speed it is perpendicular to the velocity outright.
The speedometer of a car reports one number and cannot, on its own, say whether the car is accelerating. Acceleration is the rate of change of the velocity vector, and a vector changes whenever either its length or its direction changes. A car at a steady sixty round a roundabout has constant speed and a large acceleration pointing into the circle; a car braking in a straight line has changing speed and an acceleration pointing backwards along the road.
Put algebraically: $\dfrac{d}{dt}|\mathbf{v}|$ is a scalar and $\mathbf{a}$ is a vector, and the first is only the component of the second along the direction of travel. Whenever a question says constant speed, translate it at once into $|\mathbf{v}|$ is constant, and then remember what constant length forces: $\mathbf{v} \cdot \mathbf{a} = 0$, so the acceleration is entirely sideways.
Let $\mathbf{r}(t) = \langle t^{2},\, 3t,\, t^{3} \rangle$. Differentiating once, $\mathbf{v}(t) = \langle 2t,\, 3,\, 3t^{2} \rangle$.
Velocity is the first derivative.
Differentiating again, $\mathbf{a}(t) = \langle 2,\, 0,\, 6t \rangle$. At $t = 1$ these are $\mathbf{v} = \langle 2, 3, 3 \rangle$ and $\mathbf{a} = \langle 2, 0, 6 \rangle$.
Acceleration is the second derivative.
The speed at $t = 1$ is $\sqrt{4 + 9 + 9} = \sqrt{22}$, a single number with no direction in it. Note that $\mathbf{v}$ and $\mathbf{a}$ point in visibly different directions, which is normal and is what makes the next two lessons necessary.
Speed is a length; velocity is an arrow.
A probe has $\mathbf{a}(t) = \langle 0, 2, 0 \rangle$ with $\mathbf{v}(0) = \langle 1, 0, 4 \rangle$ and $\mathbf{r}(0) = \langle 0, 0, 0 \rangle$. Integrating once: $\mathbf{v}(t) = \langle 0, 2t, 0 \rangle + \mathbf{C}_1$.
One integration, one unknown constant vector.
At $t = 0$ this must equal $\langle 1, 0, 4 \rangle$, so $\mathbf{C}_1 = \langle 1, 0, 4 \rangle$ and $\mathbf{v}(t) = \langle 1,\, 2t,\, 4 \rangle$.
The initial velocity fixes it at once.
Integrating again and using $\mathbf{r}(0) = \mathbf{0}$ gives $\mathbf{r}(t) = \langle t,\, t^{2},\, 4t \rangle$. Two integrations, two conditions, each used the moment its constant appeared — and the answer can be checked by differentiating it twice.
Fix each constant as soon as it appears.
Take $\mathbf{v}_0 = \langle 10, 0, 10 \rangle$, $\mathbf{r}_0 = \mathbf{0}$ and $\mathbf{a} = \langle 0,0,-10 \rangle$, so $\mathbf{r}(t) = \langle 10t,\; 0,\; 10t - 5t^{2} \rangle$.
The constant-acceleration formula, one component at a time.
It lands when the third component returns to zero: $10t - 5t^{2} = 0$, so $t(10 - 5t) = 0$ and the landing time is $t = 2$.
At $t = 2$ the first component is $20$, so it travels twenty metres. Note that the landing time came entirely from the vertical component and the distance entirely from the horizontal one — the two halves of the problem met only through the shared clock.
A particle has $\mathbf{r}(t) = \langle t^{2},\; t,\; t^{3} \rangle$. Fill in its velocity and acceleration at time $t = 1$, component by component.
| Velocity | Acceleration | |
|---|---|---|
| First component | ||
| Second component | ||
| Third component |
A probe's acceleration is known for every $t$, together with its velocity and position at $t = 0$. Put the steps that recover its position at $t = 4$ in order.
Number the steps in order (write the number in the box):
A stone is thrown so that $\mathbf{r}(t) = \langle 4t,\; 0,\; t - t^{2} \rangle$. At what time is it moving horizontally?
Answer:
A particle has acceleration $\mathbf{a}(t) = \langle 0, 0, -3 \rangle$ and velocity $\mathbf{v}(0) = \langle 6, 0, 5 \rangle$. What is the third component of its velocity at $t = 3$?
Answer:
A particle has $\mathbf{r}(t) = \langle 4t^{2},\; -4t,\; 2t^{3} \rangle$. Fill in its velocity and acceleration at time $t = 3$, component by component.
| Velocity | Acceleration | |
|---|---|---|
| First component | ||
| Second component | ||
| Third component |
A car rounds a bend with its speedometer fixed at $46$ kilometres per hour. Is it accelerating?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A probe's acceleration is known for every $t$, together with its velocity and position at $t = 0$. Put the steps that recover its position at $t = 5$ in order.
Number the steps in order (write the number in the box):
You can move in both directions between position, velocity and acceleration, and you can say what the speedometer does and does not tell you about acceleration. Next: measuring the length of the road itself, which does not depend on how fast it was driven.
10. Your turn: how far does a stone thrown at forty-five degrees travel?, step 3