MA High School Introductory Physics MCAS Module 1 kinematics and motion: a complete overview of scalars and vectors, velocity and acceleration, motion graphs, the kinematic equations, free fall, and projectiles

What Module 1 actually demands

Module 1 is about describing motion precisely, the foundation for everything that follows. Under the Massachusetts STE framework it sits in the Motion and Forces reporting category, the largest on the test at about half the points. The standards are written around the practices of using mathematics and computational thinking and analyzing and interpreting data, so the MCAS tests this module with calculations, motion graphs, and short scenarios far more than with bare recall. The defining tool is the reference sheet, which hands you the motion equations and leaves you to choose and apply them.

This guide ties together the matching dot-point pages, each with its own practice questions: scalars, vectors, and units, displacement, velocity, and acceleration, graphs of motion, the kinematic equations, free fall, and projectile and two-dimensional motion.

Scalars, vectors, and units

Every motion problem starts here. A scalar has size only (distance, speed, mass, time); a vector has size and direction (displacement, velocity, acceleration, force). The difference is not pedantry: a round trip has a large distance but zero displacement. The MCAS works in SI units (meters, kilograms, seconds) with metric prefixes, and the routine first step in almost every problem is to convert a value given in km/h, grams, or centimeters into SI units before substituting. The classic conversion is $72$ km/h $= 20$ m/s (divide by $3.6$).

Velocity and acceleration

Velocity is displacement over time, $v = \dfrac{d}{t}$; acceleration is the change in velocity over time, $a = \dfrac{\Delta v}{\Delta t}$. The point the MCAS hammers is that acceleration is about the change in velocity, not its size: a car at a steady $100$ km/h has high speed but zero acceleration, while a car turning a corner at constant speed is accelerating, because its direction (and so its velocity) is changing. Velocity is measured in m/s, acceleration in m/s squared.

Motion graphs

Two graph types appear, and you read them with slope and area. On a position-time graph, the slope is the velocity (steep means fast, horizontal means at rest, downward means moving backward). On a velocity-time graph, the slope is the acceleration and the area under the line is the displacement. Splitting a velocity-time graph into rectangles and triangles and adding their areas is the standard way to find total distance. Sketching a graph from a description, and describing motion from a graph, are both common tasks.

The kinematic equations

For constant acceleration, the reference sheet gives $v_f = v_i + at$ and $d = v_i t + \tfrac{1}{2}at^2$. The skill is choosing the right one: list what you know ($v_i$, $v_f$, $a$, $t$, $d$), spot hidden values ("from rest" means $v_i = 0$, "stops" means $v_f = 0$), pick the equation containing your knowns and the unknown, then substitute. Because the time-free equation is not on the sheet, a problem missing the time is solved by finding the time from the two given equations first. Signs matter: a deceleration is a negative acceleration.

Free fall

Free fall is motion under gravity alone, with acceleration $g$ (taken as $10$ m/s squared) downward. It is just the kinematic equations with $a = g$. The conceptual centerpiece is that all objects fall at the same rate regardless of mass, because $a = mg/m = g$: the heavier object's larger weight is offset by its larger mass. A dropped object starts from rest ($v_i = 0$), so $v_f = gt$ and $d = \tfrac{1}{2}gt^2$. A thrown-up object decelerates at $g$, stops at the top (where $v = 0$ but $a$ is still $g$), and falls back.

Projectile motion

A projectile's curved path is modeled as two independent motions: horizontal at constant velocity (no horizontal force) and vertical as free fall. The two are joined only by the shared time. The headline result, tested repeatedly, is that a ball thrown horizontally and a ball dropped from the same height land together, because the fall time depends only on the vertical motion. You solve a projectile problem by getting the time from the vertical drop ($d = \tfrac{1}{2}gt^2$), then the range from the horizontal motion ($d = vt$).

Check your knowledge

A mix of recall, graph, and calculation questions covering Module 1. Attempt them under timed conditions, then check against the solutions.

1. State whether each is a scalar or a vector: distance, velocity, mass, acceleration. (2 marks)
2. Convert $90$ km/h to meters per second. (1 mark)
3. A car speeds up from $5.0$ m/s to $20$ m/s in $3.0$ s. Calculate its acceleration. (2 marks)
4. On a velocity-time graph, what does the area under the line represent? (1 mark)
5. A position-time graph is horizontal. Describe the motion. (1 mark)
6. An object starts from rest and accelerates at $2.0$ m/s squared for $4.0$ s. Calculate its final velocity. (2 marks)
7. A ball is dropped from rest. Take $g = 10$ m/s squared. How far does it fall in $2.0$ s? (2 marks)
8. Explain why a heavy and a light object fall at the same rate when air resistance is ignored. (2 marks)
9. A ball is thrown horizontally from a table. What sets the time it stays in the air? (1 mark)
10. From the same height, one ball is dropped and one is thrown horizontally at the same instant. Which lands first? (1 mark)