AP Syllabus focus:
‘Apply differentiation techniques to solve rectilinear motion problems involving position, speed, velocity, or acceleration, including finding specific values at given times and solving multi-step contextual questions.’
Straight-line motion problems use derivatives to describe how an object moves along a line. Understanding relationships among position, velocity, and acceleration allows precise analysis of motion at specific moments.
Solving Straight-Line Motion Problems
Understanding Motion Quantities
In straight-line motion, an object’s movement is modeled using a position function that describes location over time and its derivatives, which generate velocity and acceleration. These quantities allow us to determine how fast and in what direction the object moves at any given instant.
Position Function: A function describing an object’s location on a line as a function of time.
Because motion takes place along a single axis, all values indicate direction with positive or negative signs. This sign structure is essential for interpreting movement and determining when the object switches direction.
Connecting Derivatives to Motion
The velocity of an object is the derivative of its position function; the acceleration is the derivative of velocity. These connected rates make it possible to solve multi-step contextual questions, such as finding when an object stops, when it changes direction, or how its speed is evolving over time.
= Velocity; rate of change of position
= Position function
= Acceleration; rate of change of velocity
These formulas help determine how motion behaves at particular moments and allow extraction of important physical information from functions, graphs, or descriptions.
Essential Tasks in Straight-Line Motion Problems
Motion problems often require determining a position, velocity, or acceleration value at a specific time, or analyzing how those values interact. To achieve this, students must be comfortable moving between representations.
Common tasks include:
Finding position at a given time using the position function.
Computing instantaneous velocity using derivatives.
Determining speed, which is the absolute value of velocity.
Evaluating acceleration to understand how motion is changing.
Identifying intervals of speeding up or slowing down by comparing the signs of velocity and acceleration.
Solving multi-step problems in which earlier results become inputs for later steps.
Speed and Its Interpretation
Although closely related to velocity, speed reflects only magnitude, not direction. Therefore, analyzing speed requires carefully interpreting the velocity’s sign while using its absolute value.
Speed: The magnitude of velocity, representing how fast an object is moving regardless of direction.
Once speed is found, students can infer whether the object is increasing or decreasing its rate of motion depending on the interplay between velocity and acceleration.
Determining Direction and Changes in Motion
Key moments in straight-line motion arise when the object stops or reverses direction. These transitions occur when velocity equals zero. Solving these conditions often involves identifying roots of the derivative of the position function, then evaluating acceleration or surrounding values to understand the object's behavior.
Students should be able to:
Interpret the meaning of a zero velocity value.
Evaluate acceleration at the same instant to determine whether the object is slowing, speeding, or reversing.
Use sign charts or logical reasoning to analyze intervals of motion.
These interpretive skills form the basis of multi-step contextual problem solving.
Position, velocity, and acceleration graphs for vertical one-dimensional motion under constant acceleration. The velocity graph crosses zero at the peak of motion, while acceleration remains constant. The context involves a vertically thrown rock, slightly more specific than required, but the motion relationships match straight-line calculus analysis. Source.
Multi-Step Problem Structure
Straight-line motion problems frequently require sequential reasoning. A single problem might ask for position at a time, then the corresponding velocity, then whether the object is speeding up or slowing down. To manage these tasks effectively, students should approach them systematically.
A useful structure includes:
Reading the scenario carefully to identify which motion quantities are involved.
Determining whether differentiation or evaluation is required first.
Tracking units to maintain consistency and accuracy.
Recognizing intermediate results that must be carried forward.
Solving such layered problems strengthens conceptual understanding and prepares students for more complex calculus applications.
Using Provided Information Strategically
Some motion problems give a graph or table instead of an explicit formula. In these cases, students may need to approximate derivatives, estimate values, or infer direction of motion based on slopes and data changes.
Effective approaches include:
Using the slope of a tangent line on a graph to approximate instantaneous velocity.
Reading values from tables to compute average rates over small intervals.
Observing how signs of estimated velocities indicate direction.
Such methods extend derivative understanding beyond algebraic contexts and reinforce the syllabus’s emphasis on interpreting motion in various representations.
Displacement–time and velocity–time plots for constant-acceleration motion. The increasing slope in the displacement graph corresponds to the linear velocity graph. The page also includes derived formulas beyond the syllabus, but the image itself focuses on graphical relationships essential to straight-line motion. Source.
FAQ
If a problem refers to a value “at time t = …”, this almost always indicates an instantaneous quantity, which requires differentiation.
If it refers to change “over an interval”, it typically indicates an average rate.
Clues include:
• Instantaneous values rely on derivatives.
• Average values rely on differences in position or velocity over time.
When in doubt, check whether the information describes a single moment or a span of time.
Direction changes occur when the velocity changes sign, which corresponds to a position graph having a turning point.
Look for:
• A clear maximum or minimum in the position curve.
• The slope changing from positive to negative, or vice versa.
• Points where the tangent is horizontal.
These features indicate zero velocity and potential reversal of motion.
Speed is the magnitude of velocity and ignores direction, making it useful for understanding how rapidly motion occurs regardless of positive or negative movement.
This distinction matters when:
• Velocity is negative but speed should still be treated as positive.
• Determining whether an object is speeding up or slowing down, which depends on comparing signs of velocity and acceleration rather than magnitudes alone.
Speed increases when velocity and acceleration share the same sign, and decreases when they differ. Velocity graphs show both position of the curve relative to the time axis and the trend of the slope.
On the graph:
• Above the axis with an increasing slope indicates speeding up.
• Above the axis with a decreasing slope indicates slowing down.
• Below the axis with decreasing slope indicates speeding up (in the negative direction).
• Below the axis with increasing slope indicates slowing down.
Break the problem into identifiable parts based on what quantity is required: position, velocity, speed, or acceleration.
Useful strategies include:
• Differentiating or evaluating the given function before reading ahead.
• Clearly writing intermediate quantities so they can be reused.
• Checking unit consistency at each stage.
• Looking for points where motion changes behaviour, such as zero velocity or sign changes.
Organising the logic early prevents wasted time and unnecessary recalculation.
Practice Questions
Question 1 (1–3 marks)
A particle moves along a straight line with position given by s(t) = t² − 4t, where s is measured in metres and t in seconds.
Find the velocity of the particle at t = 3.
Question 1
• Correct differentiation of s(t) to obtain v(t) = 2t − 4. (1 mark)
• Correct substitution t = 3 into v(t). (1 mark)
• Correct final answer: velocity = 2 m/s. (1 mark)
Question 2 (4–6 marks)
An object moves along a horizontal line. Its velocity v(t), measured in metres per second, is given by v(t) = 6t − t² for t ≥ 0.
(a) Determine the time at which the object is at rest.
(b) Determine the object’s acceleration at this time.
(c) Determine whether the object is speeding up or slowing down at this instant, justifying your answer using the signs of velocity and acceleration.
Question 2
(a)
• Correct method: set v(t) = 0. (1 mark)
• Correct solution: 6t − t² = 0 leading to t = 6. (1 mark)
(b)
• Correct differentiation of v(t) to obtain a(t) = 6 − 2t. (1 mark)
• Correct substitution t = 6 to give a(6) = -6 m/s². (1 mark)
(c)
• Correct interpretation of signs: velocity is zero and acceleration is negative. (1 mark)
• Correct conclusion: the object is slowing down at that instant (or beginning to move in the negative direction from rest, which is consistent with slowing relative to previous positive motion). (1 mark)
