B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Position, Velocity and Acceleration as Derivatives

B2.1About 70 minutes

Learning goals

I can find velocity and acceleration functions by differentiating a position function.
I can interpret the sign and units of position, velocity and acceleration.
I can find when an object is at rest and in which direction it is moving.
I can solve vertical-motion and braking problems with derivatives.

Words to know

position s(t)
Where an object is at time t, measured from a chosen zero point along a line, with a chosen positive direction.

displacement
The change in position, s(b) − s(a). It can be positive, negative or zero.

velocity
The instantaneous rate of change of position: v(t) = s′(t). Its sign gives the direction of motion.

speed
The size of the velocity, |v(t)|. It is never negative.

acceleration
The instantaneous rate of change of velocity: a(t) = v′(t) = s″(t).

at rest
Having velocity 0 at that instant.

Your speedometer is a derivative

When a car's speedometer reads 60 km/h, nobody is claiming the car will cover exactly 60 km in the next hour. The needle shows how fast the car is going right now. That is an instantaneous rate of change, which is exactly what a derivative measures.

Remember from Grade 11: in Physics (SPH3U, the unit Kinematics) you found velocity as the slope of a position-time graph and acceleration as the slope of a velocity-time graph, and you used the equations of uniformly accelerated motion. In MHF4U (Grade 12 Advanced Functions, taken before or alongside this course), in the unit Rates of Change, you estimated instantaneous rates with secants and the difference quotient. Calculus now gives you exact slopes at every instant, for motion that is not uniformly accelerated too.

Position, velocity and acceleration

Picture an object moving along a straight line (a track, a road, a vertical line for a thrown ball). Choose a zero point and a positive direction.

Position s(t) is where the object is at time t, measured from the zero point. Positive means on the positive side.
Displacement from t = a to t = b is the change in position: s(b) − s(a).
Velocity is the instantaneous rate of change of position: v(t) = s′(t). In Leibniz notation, v = ds/dt.
Acceleration is the instantaneous rate of change of velocity: a(t) = v′(t) = s″(t). So acceleration is the second derivative of position.
Speed is the size of the velocity, |v(t)|. Speed is never negative.

If s is in metres and t in seconds, then v is in m/s (metres per second) and a is in m/s² (metres per second, per second). The units always follow the derivative: the unit of the top quantity divided by the unit of time.

What the signs tell you

v(t) > 0: the object is moving in the positive direction.
v(t) < 0: it is moving in the negative direction.
v(t) = 0: it is momentarily at rest. If v changes sign there, the object turns around.
a(t) > 0: the velocity is increasing. a(t) < 0: the velocity is decreasing.

Careful: a negative acceleration does not always mean slowing down. You will sort out speeding up and slowing down in the next lesson.

Worked example

Example 1: A particle on a line

A particle moves along a line so that its position is s(t) = t³ − 6t² + 9t metres, t ≥ 0 seconds. (a) Find v(t) and a(t). (b) Find the position, velocity and acceleration at t = 2. (c) When is the particle at rest? (d) When is it moving in the positive direction?

(a) Differentiate with the power rule: v(t) = s′(t) = 3t² − 12t + 9. Differentiate again: a(t) = v′(t) = 6t − 12.
(b) s(2) = 8 − 24 + 18 = 2 m. v(2) = 12 − 24 + 9 = −3 m/s. a(2) = 12 − 12 = 0 m/s².
(c) At rest means v(t) = 0: 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3) = 0, so t = 1 s or t = 3 s.
(d) Make a sign chart for v(t) = 3(t − 1)(t − 3). For 0 ≤ t < 1 both factors are negative, so v > 0. For 1 < t < 3, v < 0. For t > 3, v > 0.
So the particle moves in the positive direction for 0 ≤ t < 1 and t > 3, and in the negative direction for 1 < t < 3.

s(2) = 2 m, v(2) = −3 m/s (moving in the negative direction at 3 m/s), a(2) = 0. At rest at t = 1 s and t = 3 s. Positive direction on [0, 1) and (3, ∞).

The path of the particle in Example 1, drawn above a number line. It starts at s = 0, moves right to s = 4, turns around at t = 1, comes back to s = 0 at t = 3, then heads right again (it is back at s = 4 when t = 4).

The three graphs share one time axis. Where the s graph has a horizontal tangent (t = 1 and t = 3), v is zero. Where the v graph has its lowest point (t = 2), a is zero. Each graph is the slope of the one above it.

Worked example

Example 2: A ball thrown straight up

A ball is thrown straight up from a deck 1.5 m above the ground. Its height is h(t) = −4.9t² + 14.7t + 1.5 metres after t seconds. (a) Find the velocity and acceleration. (b) When does the ball reach its maximum height, and how high is it? (c) How fast is it going when it hits the ground?

(a) v(t) = h′(t) = −9.8t + 14.7 m/s and a(t) = v′(t) = −9.8 m/s². The acceleration is constant: it is the acceleration due to gravity, pointing down (up is positive here).
(b) At the top the ball is momentarily at rest, so v(t) = 0: −9.8t + 14.7 = 0 gives t = 1.5 s. Then h(1.5) = −4.9(2.25) + 14.7(1.5) + 1.5 = −11.025 + 22.05 + 1.5 = 12.525 m.
(c) The ball hits the ground when h(t) = 0: −4.9t² + 14.7t + 1.5 = 0. The quadratic formula gives t = [−14.7 − √(14.7² + 4(4.9)(1.5))] ÷ (2(−4.9)) = (14.7 + √245.49) ÷ 9.8 ≈ (14.7 + 15.668) ÷ 9.8 ≈ 3.099 s. (The other root is negative, so we reject it.)
v(3.099) ≈ −9.8(3.099) + 14.7 ≈ −15.67 m/s. The negative sign means the ball is moving down.

Maximum height about 12.5 m at t = 1.5 s. It lands after about 3.10 s with velocity about −15.7 m/s, so its speed is about 15.7 m/s.

Worked example

Example 3: A train pulling into a station

As a GO train brakes into a station, the distance it has travelled since the brakes went on is s(t) = 24t − 0.6t² metres, for 0 ≤ t ≤ 20 seconds. How fast was it going when braking began, when does it stop, and how far does it travel while braking?

v(t) = s′(t) = 24 − 1.2t. At t = 0, v = 24 m/s (that is 24 × 3.6 = 86.4 km/h).
It stops when v(t) = 0: 24 − 1.2t = 0, so t = 20 s. (That is why the model is only used for 0 ≤ t ≤ 20.)
Braking distance = s(20) − s(0) = 24(20) − 0.6(400) − 0 = 480 − 240 = 240 m.
a(t) = −1.2 m/s² for the whole time: a constant deceleration.

Initial velocity 24 m/s (86.4 km/h); it stops after 20 s, having travelled 240 m.

Worked example

Example 4: A mass bouncing on a spring

A mass on a spring moves up and down so that its position (in cm, above its resting point) is s(t) = 5 cos(πt), where t is in seconds. Find v(t) and a(t), and show that the acceleration is always proportional to the position.

Use the derivatives from Derivatives of Exponential and Sinusoidal Functions and the chain rule: d/dt cos(πt) = −π sin(πt).
v(t) = s′(t) = −5π sin(πt) cm/s.
a(t) = v′(t) = −5π · π cos(πt) = −5π² cos(πt) cm/s².
Compare with s(t) = 5 cos(πt): a(t) = −π² × 5 cos(πt) = −π² s(t).

a(t) = −π² s(t). The acceleration always points back toward the resting point and is biggest when the mass is farthest away. Physicists call this simple harmonic motion; you will see it again in university physics.

Just for fun

Why did the particle refuse to tell anyone where it was going? It said its velocity was a personal derivative. (Acceleration, being a second derivative, was even more private.)

Hands-on activity

Make your own position-time data

You need: a long hallway or driveway (no traffic), a tape measure, masking tape or sidewalk chalk, a phone stopwatch, a helper, and paper.

Mark a start line and then a mark every 1 m for 8 m.
Your helper starts the stopwatch and calls out each second ("one, two, three..."). Starting from rest, walk slowly at first and then faster and faster along the line. At each call, notice which mark you are passing and say it out loud; the helper writes it down.
Make a table of t (s) and s (m). Estimate the velocity in each one-second interval with Δs/Δt (an average rate, as in MHF4U).
Fit a model: try s(t) = kt². Choose k so that the model matches your last data point, then compare the model's derivative v(t) = 2kt with your estimates.
Repeat, walking at a steady pace. What do your velocity estimates look like now? What is the acceleration?

What to look for: when you speed up, the Δs/Δt values grow and are close to the straight line v = 2kt; the acceleration 2k is constant. At a steady pace the Δs/Δt values are about the same, so the acceleration is about 0. If you have a free phone sensor app such as phyphox, its accelerometer can show the same pattern.

Your turn

s(t) = 2t² − 8t + 3 (metres, seconds). Find v(t) and a(t).

For s(t) = 2t² − 8t + 3, when is the object at rest, and where is it then?

s(t) = t³ − 3t² (m, s). Find the velocity and acceleration at t = 3.

A stone is dropped from a bridge 45 m above a river. Its height is h(t) = 45 − 4.9t². Find its velocity after 2 s.

For the stone in the previous question, how long does it take to reach the water, and how fast is it going then?

s(t) = t⁴ − 4t³ (m, s), t ≥ 0. Find all times when the object is at rest.

A cyclist's position is s(t) = 0.2t³ − 1.5t² + 6t metres. Show that the cyclist never stops (never has v = 0).

More practice: warm-up

If s is measured in metres and t in seconds, what are the units of s′(t)?

Which derivative of position is acceleration?

s(t) = 5t + 2. Find v(t).

v(t) = 3t² − 2t. Find a(t).

A velocity is −4 m/s. What is the speed?

More practice: on your own

s(t) = −t³ + 6t² − 9t + 10 (m, s), t ≥ 0. When is the object moving in the positive direction?

For s(t) = t³ − 9t² + 24t, find the acceleration at each time the object is at rest.

A ball's height is h(t) = −4.9t² + 19.6t + 2 (m). Find the maximum height.

A car's position is s(t) = 30t − 0.5t² m during braking. How far does it travel before stopping?

s(t) = 4 sin(2t) metres. Find v(0) and a(π/4).

Challenge

An object's position is s(t) = t + 4/t for t > 0 (m, s). Find when it is at rest and its acceleration then.

A ball thrown upward from the ground has h(t) = v₀t − 4.9t². It reaches a maximum height of 20 m. Find its initial velocity v₀.

Show that for any position function s(t) = At² + Bt + C, the average velocity over [t₁, t₂] equals the instantaneous velocity at the midpoint of the interval.

Check your understanding

Quiz

An object's position is s(t) = t³ − 12t (m, s). What is its velocity at t = 3?

Which statement is always true?

If s is in metres and t is in seconds, what are the units of s″(t)?

For s(t) = t² − 6t + 5, when is the object at rest?

A ball's height is h(t) = −4.9t² + 9.8t + 1. What is its maximum height?

For s(t) = 5 cos(πt) cm, which expression is the acceleration?

For parents and teachers

Big idea (B2.1): velocity is the derivative of position, and acceleration is the derivative of velocity (the second derivative of position). Signs carry meaning: the sign of v tells direction, v = 0 means momentarily at rest.

How to help: ask your teen to explain what each number in an answer means in words, with units ("at 2 seconds the particle is 2 m to the right of the start, moving left at 3 m/s"). If they took Grade 11 physics, ask how the slope of a position-time graph connects to today's work.

Common difficulties:

Confusing position with velocity (answering "when is s = 0?" when asked "when is it at rest?").
Treating speed and velocity as the same thing. Speed is |v|.
Forgetting units, or writing m/s for acceleration instead of m/s².
Using a model outside its domain (the braking train after it has stopped).