Derivatives translate directly into physics: the derivative of position is velocity, and the derivative of velocity is acceleration. This section teaches you to read velocity graphs, distinguish speed from velocity, and solve projectile-motion problems by connecting the algebra of derivatives to the behaviour of moving objects.
Difficulty: Intermediate
Prerequisites: Basic differentiation (power rule), the chain rule (Sec. 2.5), and familiarity with reading function graphs.
Section 2.7 is where calculus meets physics. You have spent weeks learning how to compute derivatives; now you apply them to describe how objects move. Position, velocity and acceleration form a derivative chain: each is the derivative of the one before. This material is a gateway to related rates and optimisation, and it appears on virtually every Calculus I final. If you can read a velocity graph and set up a projectile problem, you are well placed for what comes next.
Position function, s(t) or h(t)
Gives the location of a particle on a line (or the height of an object) at time t. Units are typically metres.
Velocity, v(t)
The derivative of the position function: v(t) = s'(t). Velocity has a sign: positive means the particle moves in the positive direction (often rightward or upward), negative means it moves in the negative direction. Think of it as speed with a direction attached.
Speed, |v(t)|
The absolute value of velocity. Speed is always non-negative. It tells you how fast the particle is moving, with no information about direction.
Acceleration, a(t)
The derivative of velocity (and the second derivative of position): a(t) = v'(t) = s''(t). It tells you how the velocity is changing. Positive acceleration does not necessarily mean the particle is speeding up.
Displacement
The net change in position over a time interval: s(b) − s(a). It can be zero even if the particle moved, because motion in opposite directions cancels. Think of it as "where you end up relative to where you started."
Total distance travelled
The sum of all distances covered regardless of direction. It is always greater than or equal to the magnitude of displacement. Think of it as what the odometer reads.
Speed tells you how fast the particle is moving. It is the absolute value of velocity and is always ≥ 0. When someone says "the particle has a speed of 5 m/s," you know the particle is moving at 5 m/s but you do not know the direction.
Velocity tells you how fast and in which direction. When someone says "the particle has a velocity of 5 m/s," the particle is moving at 5 m/s in the positive direction. A velocity of −5 m/s means 5 m/s in the negative direction.
The relationship:
speed = |velocity|
If v(t) > 0, the particle moves in the positive direction
If v(t) < 0, the particle moves in the negative direction
If v(t) = 0, the particle is momentarily at rest
These are not the same thing, and confusing them is one of the most common errors on exams.
Positive acceleration means a(t) > 0, i.e. the velocity is increasing. If the particle is moving in the negative direction (v < 0) and has positive acceleration, the particle is actually slowing down (its speed is decreasing).
Speeding up means the speed (|v(t)|) is increasing. This happens when velocity and acceleration have the same sign:
v > 0 and a > 0: moving forward, getting faster. Speeding up.
v < 0 and a < 0: moving backward, getting faster (in the negative direction). Speeding up.
v > 0 and a < 0: moving forward, slowing down.
v < 0 and a > 0: moving backward, slowing down.
The rule: the particle speeds up when v(t) and a(t) share the same sign. It slows down when they have opposite signs.
Displacement is the net effect of motion on position. If a particle starts at x = 2 and ends at x = 5, its displacement is 3, regardless of what happened in between. Displacement can be negative (if the particle ends to the left of where it started) or zero (if it returns to its starting point).
Total distance travelled is the sum of all distances, counting every stretch of motion positively. If the particle goes from x = 2 to x = 8 and then back to x = 5, the displacement is 3 but the total distance is 6 + 3 = 9.
How to compute each from a velocity function:
Displacement from t = a to t = b: the integral of v(t) from a to b (this is a Calc II technique, but the concept matters now)
Total distance from t = a to t = b: the integral of |v(t)| from a to b
For now, the key is the conceptual distinction. Exams will ask you to explain the difference in words.
The recitation gives a graph of v(t) for a particle on a horizontal line, with t from 0 to 10 and v(t) ranging from −3 to 3 m/s. Here is how to extract every piece of information the exam will ask for.
Positive/negative direction
The particle moves in the positive direction when v(t) > 0. Read the intervals where the graph sits above the t-axis. From the recitation graph: (0, 5).
The particle moves in the negative direction when v(t) < 0. Read the intervals below the t-axis: (7, 10).
Maximum speed vs maximum velocity
Maximum speed = max |v(t)|. Look for the point farthest from the t-axis in either direction. From the graph, the lowest point is v = −3, so max speed = |−3| = 3 m/s.
Maximum velocity = max v(t). Look for the highest point on the graph. From the graph, max velocity = 2 m/s.
These are different numbers whenever the graph dips below the axis farther than it rises above it.
When is the particle motionless?
When v(t) = 0, i.e. where the graph crosses or touches the t-axis. From the recitation graph: the interval [5, 7).
Acceleration from a velocity graph
Acceleration = v'(t) = the slope of the velocity graph.
Acceleration is positive where the graph slopes upward: (0, 2) ∪ (8, 10)
Acceleration is negative where the graph slopes downward: (4, 5) ∪ (7, 8)
When is the speed increasing/decreasing?
Speed increases when v and a share the same sign:
Increasing: (0, 2) ∪ [7, 8) (both positive in the first interval; both negative in the second)
Decreasing: (4, 5) ∪ (8, 10)
Snapshot at a specific time (t = 9)
At t = 9 on the recitation graph, v(9) < 0 (the graph is below the axis), so the particle is moving in the negative direction. The slope of v at t = 9 is positive (the graph is rising), so acceleration is positive. Because v < 0 and a > 0, the particle is slowing down.
A ball is thrown vertically upward with initial velocity 10 m/s. Its height is h(t) = −at² + 10t + 1.
Part (a): finding the constant a, given that maximum height is reached at t = 5
At maximum height the velocity is zero. Differentiate: v(t) = h'(t) = −2at + 10. Set v(5) = 0:
−10a + 10 = 0, so a = 1.
The height function is therefore h(t) = −t² + 10t + 1.
Part (b): maximum height
Plug t = 5 into h(t): h(5) = −25 + 50 + 1 = 26 metres.
Part (c): when does the ball hit the ground?
Set h(t) = 0: −t² + 10t + 1 = 0, or t² − 10t − 1 = 0.
Using the quadratic formula: t = (10 ± √104) / 2 = 5 ± √26.
Since t must be positive and greater than the launch time, the ball lands at t = 5 + √26 seconds (approximately 10.1 s).
Part (d): speed at landing
Evaluate v(t) at t = 5 + √26: v = −2(5 + √26) + 10 = −2√26 m/s.
The velocity is negative (the ball is falling), so the speed is |v| = 2√26 m/s (approximately 10.2 m/s).
General pattern for projectile problems
Start from the position function h(t)
Differentiate to get v(t)
Set v(t) = 0 to find the time of maximum height
Plug that time back into h(t) for the maximum height
Set h(t) = 0 and solve for t to find when the object hits the ground
Evaluate |v(t)| at landing time for the speed at impact
Students often think "positive acceleration" and "speeding up" mean the same thing. They do not. A particle moving in the negative direction with positive acceleration is slowing down.
Students confuse "the particle is at rest" with "the particle is at the origin." At rest means v(t) = 0; at the origin means s(t) = 0. These are independent conditions.
Students sometimes read maximum speed off a velocity graph by looking only at the highest point. Maximum speed is the point farthest from the t-axis in either direction, so a deep negative trough can give a higher speed than a shallow positive peak.
In projectile problems, students forget that a negative velocity at landing is expected (the object is falling) and report a negative speed. Speed is always |v|, never negative.
⚠️ Expect a velocity-graph problem on the exam. You will be asked some combination of: direction of motion, when the particle is at rest, when it is speeding up vs slowing down, and the maximum speed.
⚠️ The distinction between speed and velocity is a favourite short-answer or true/false question.
⚠️ Projectile-motion problems test whether you can connect derivatives to physical quantities. The setup is always: differentiate h(t), set v = 0, solve for t, plug back in.
⚠️ Displacement vs total distance is frequently tested as a conceptual question. Be ready to explain the difference in one or two sentences.
True or False: A particle with positive acceleration is always speeding up.
Fill in the blank: Speed is the ________ of velocity.
True or False: Displacement can be zero even if the particle moved.
On a velocity graph, the acceleration at time t equals the ________ of the graph at that point.
True or False: Maximum speed and maximum velocity are always the same number.
Answers: 1. False (it speeds up only if velocity is also positive). 2. absolute value. 3. True (the particle can return to its starting position). 4. slope. 5. False (maximum speed = max |v(t)|, which can occur at a negative velocity value).
Q: A particle moves along a horizontal line with velocity v(t) shown on a graph. v(t) > 0 on (0, 5), v(t) = 0 on [5, 7), and v(t) < 0 on (7, 10). On what intervals is the particle moving in the negative direction?
A: The particle moves in the negative direction when v(t) < 0, which is on the interval (7, 10).
Q: On the same graph, the highest point is v = 2 and the lowest point is v = −3. What is the maximum speed? What is the maximum velocity?
A: Maximum speed = max |v(t)| = |−3| = 3 m/s. Maximum velocity = max v(t) = 2 m/s.
Q: A ball is thrown upward with h(t) = −t² + 10t + 1. When does it reach maximum height, and what is that height?
A: v(t) = h'(t) = −2t + 10. Set v(t) = 0: t = 5. Maximum height = h(5) = −25 + 50 + 1 = 26 metres.
Q: For the same ball, when does it hit the ground?
A: Set h(t) = 0: −t² + 10t + 1 = 0, so t² − 10t − 1 = 0. By the quadratic formula, t = 5 ± √26. Since t must be positive and after launch, the ball lands at t = 5 + √26 ≈ 10.1 seconds.
Q: What is the ball's speed at the moment it hits the ground?
A: v(5 + √26) = −2(5 + √26) + 10 = −2√26. Speed = |−2√26| = 2√26 ≈ 10.2 m/s.
Q: Explain, in one sentence, the difference between displacement and total distance travelled.
A: Displacement is the net change in position (final minus initial, which can be negative or zero), while total distance travelled is the sum of all distances covered regardless of direction (always non-negative).
Particle motion is your first encounter with the idea that derivatives describe rates of change in a physical system. This same reasoning drives related-rates problems (Sec. 3.9), where you differentiate equations connecting multiple changing quantities. Optimisation problems (Ch. 4) use the same "set the derivative to zero" technique you used here to find maximum height. In Calculus II, integration will let you go the other direction: from velocity back to position, and from acceleration back to velocity.
Velocity, speed, acceleration, position function, displacement, total distance travelled, particle motion, velocity graph, speeding up, slowing down, projectile motion, quadratic formula, tangent line, maximum height, time of flight, MTH 132, Calculus I, MSU, Michigan State University, Sec. 2.7, rectilinear motion, kinematics