Difficulty: Introductory to Intermediate | Prerequisites: Algebra, basic trigonometry
One-dimensional kinematics is the backbone of the first half of PHYSICS 114. Every topic that follows (projectile motion, forces, energy) builds on your ability to relate position, velocity, and acceleration using graphs and equations. If the kinematic equations and graph reading feel automatic, the rest of the course is significantly easier. This material assumes you are comfortable with signed numbers and basic algebra.
Kinematics describes how things move without worrying about why. Position, velocity, and acceleration are related by a small set of equations that work whenever acceleration is constant. The slope of a position-time graph gives velocity; the slope of a velocity-time graph gives acceleration; and the area under a velocity-time graph gives displacement.
Position (x or y)
Where an object is along a chosen axis, measured from an origin. It is a signed number: positive in one direction, negative in the other.
Displacement (delta x)
The change in position: delta x = x_final - x_initial. It is a vector (has direction). Think of it as "how far and which way did the object end up from where it started."
Velocity (v)
The rate of change of position with respect to time. Average velocity = delta x / delta t. It has a sign indicating direction.
Instantaneous velocity
The velocity at a single moment in time. On a position-time graph, it equals the slope of the tangent line at that point. In simple terms, it is what the speedometer reads (with a sign for direction).
Acceleration (a)
The rate of change of velocity with respect to time. Average acceleration = delta v / delta t. Positive acceleration does not always mean speeding up; it means velocity is becoming more positive.
Free fall
Motion under gravity alone, with a = -g = -9.80 m/s^2 (taking upward as positive). The acceleration is constant and directed downward regardless of whether the object is moving up or down.
Kinematic equations
A set of four equations relating position, velocity, acceleration, and time for constant-acceleration motion. Each equation leaves out one of the five variables (x, v, v0, a, t), so you pick the equation that matches the variables you know and the one you want.
The slope of a position-time graph at any point equals the instantaneous velocity at that time.
A straight section means constant velocity. A curved section means the velocity is changing (the object is accelerating).
A horizontal tangent (slope = 0) means the object is momentarily at rest.
For Question 3: at t = 2 s, the position-time curve has a tangent line that is horizontal (the curve reaches a minimum around t = 2 s and starts turning). Slope = 0, so instantaneous velocity = 0 m/s. Answer: D) 0 m/s.
The slope of a v-t graph gives acceleration.
The area under a v-t graph (between the curve and the time axis) gives displacement.
A constant positive value on a v-t graph means constant positive velocity (no acceleration).
A line sloping upward means positive acceleration; sloping downward means negative acceleration (which could be slowing down or speeding up in the negative direction).
A car goes from x = 50 m to x = 10 m: displacement is negative (it moves in the negative direction).
It has constant positive acceleration: velocity is becoming more positive over time.
So the car starts with a negative velocity and that velocity increases (becomes less negative, passes through zero, or reaches a less-negative value). The v-t graph should show a line starting below zero and rising with constant slope.
Answer: D (a line starting negative and increasing linearly).
Distance travelled = area under the v-t curve from t = 0 to t = 5 s.
Andy's graph: constant velocity of 40 m/s for 5 s, area = 200 m.
Betty's graph: linearly increasing from 0 to 40 m/s over 5 s, area = 0.5 x 40 x 5 = 100 m.
Carl's graph: velocity increases in a concave-up curve from 0 to 40 m/s, area is less than the triangle (less than 100 m).
Answer: A) Andy.
At maximum height, velocity = 0. Use v^2 = v0^2 + 2a(delta y).
0 = v0^2 + 2(-9.80)(1.10). Solve: v0^2 = 2(9.80)(1.10) = 21.56, so v0 = 4.64 m/s.
Answer: B) 4.6 m/s.
Phase 1 (engine burn): the rocket reaches y = 74.0 m with v = 30.0 m/s upward.
Phase 2 (unpowered flight): starts at y = 74.0 m, v0 = 30.0 m/s, a = -9.80 m/s^2. Find additional height until v = 0.
v^2 = v0^2 + 2a(delta y). 0 = (30.0)^2 + 2(-9.80)(delta y). delta y = 900/19.6 = 45.9 m.
Maximum height = 74.0 + 45.9 = 119.9 m, closest to 120 m.
Answer: A) 120 m.
v = v_0 + atx = x_0 + v_0 t + \frac{1}{2}at^2v^2 = v_0^2 + 2a(x - x_0)x - x_0 = \frac{1}{2}(v_0 + v)tFor free fall, replace a with -g (taking upward as positive):
a = -g = -9.80 \; \text{m/s}^2Kinematic equations describe everything from a ball tossed in the air to the braking distance of a car. Engineers use them to design roller coasters, calculate runway lengths, and determine safe following distances. Free-fall calculations are essential for anyone working with dropped or launched objects, from construction safety to sports science.
Students often confuse velocity and speed. Velocity has a sign (direction). An object moving to the left with increasing speed has a negative velocity that is becoming more negative, which means a negative acceleration.
"Positive acceleration" does not mean "speeding up." A car moving in the negative direction with positive acceleration is slowing down.
At the peak of a vertical throw, velocity is zero but acceleration is still -9.80 m/s^2. Students often think acceleration is zero at the top.
In two-phase problems (like the rocket), students sometimes apply a single kinematic equation across both phases. You cannot do this because acceleration changes between phases. Treat each phase separately, using the end conditions of phase 1 as the initial conditions of phase 2.
Graph-reading questions appear on virtually every exam. Practise identifying slope and area on both x-t and v-t graphs.
Free-fall problems are a staple. Remember to define your sign convention (usually up = positive) and stick with it.
Two-phase problems (engine burn then free flight, or car accelerating then braking) are a favourite exam format. The key is recognising the boundary conditions between phases.
True or false: At the highest point of a ball thrown straight up, both velocity and acceleration are zero. (False. Velocity is zero, but acceleration is -9.80 m/s^2.)
The slope of a position-time graph gives ______. (velocity)
The area under a velocity-time graph gives ______. (displacement)
True or false: If a car has positive acceleration, it must be speeding up. (False. It could be slowing down while moving in the negative direction.)
A rocket reaches 50 m with a speed of 20 m/s when its engine cuts out. What additional height does it gain? (v^2 = v0^2 + 2a(delta y), so delta y = 20^2 / (2 x 9.80) = 20.4 m)
Q: On a position-time graph, an object's curve has a horizontal tangent at t = 2 s. What is the instantaneous velocity at t = 2 s?
A: 0 m/s. The slope of the tangent line is zero, so the velocity is zero at that instant.
Q: A car moves from x = 50 m to x = 10 m with constant positive acceleration. Describe what its velocity-time graph looks like.
A: The car moves in the negative direction (displacement is negative), so its initial velocity is negative. Constant positive acceleration means the v-t graph is a straight line with a positive slope. The line starts below the time axis and rises.
Q: Three cars have the following v-t graphs over 5 seconds: Car A is constant at 40 m/s, Car B increases linearly from 0 to 40 m/s, Car C curves upward from 0 to 40 m/s (concave up). Rank by distance travelled.
A: A > B > C. The area under each curve gives displacement. A's rectangle (200 m) is largest, B's triangle (100 m) is next, and C's concave-up curve has less area than the triangle.
Q: With what speed must you leave the ground to jump 1.10 m straight up?
A: Use v^2 = v0^2 + 2a(delta y) with v = 0 at the top: v0 = sqrt(2 x 9.80 x 1.10) = sqrt(21.56) = 4.64 m/s.
Q: A toy rocket reaches 74.0 m with velocity 30.0 m/s when its engine cuts out. What is its maximum height?
A: After burnout, v0 = 30.0 m/s, a = -9.80 m/s^2. Additional height: delta y = v0^2 / (2g) = 900/19.6 = 45.9 m. Maximum height = 74.0 + 45.9 = 119.9 m, approximately 120 m.
One-dimensional kinematics extends directly into two-dimensional projectile motion, where you treat horizontal and vertical components independently, each as its own 1D problem. Newton's second law (F = ma) connects forces to the acceleration you have been calculating here. Energy methods (kinetic energy, potential energy) provide an alternative to kinematics for solving many of the same problems, often more efficiently.
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