Source: ENGR 216 Lectures, PHYS 206 Concurrent
Tags: kinematics, position, velocity, acceleration, P/V/A, kinematic equations, motion graphs, displacement, ENGR 216, PHYS 206, mechanics
Kinematics describes how objects move without worrying about why they move. The three linked quantities are position, velocity, and acceleration. Velocity is the rate of change of position; acceleration is the rate of change of velocity. Reading and interpreting motion graphs (position-time, velocity-time, acceleration-time) is a core skill tested heavily in this course.
Position (x or s)
The location of an object relative to a chosen origin, measured in metres (m). Position is a vector; in 1D problems, sign indicates direction.
Displacement (Δx)
The change in position: Δx = x_final − x_initial. Displacement is a vector (it has direction), unlike distance, which is the total path length travelled and is always positive.
Velocity (v)
The rate of change of position with respect to time: v = dx/dt. Measured in m/s. Average velocity is Δx/Δt; instantaneous velocity is the derivative of position.
Speed
The magnitude of velocity. Speed is always positive or zero; velocity can be negative (indicating direction).
Acceleration (a)
The rate of change of velocity with respect to time: a = dv/dt = d²x/dt². Measured in m/s². An object can be moving in one direction while accelerating in the opposite direction (this means it is slowing down).
Uniform (constant) acceleration
Motion where acceleration does not change over time. This is the regime where the standard kinematic equations apply directly.
Free fall
Motion under gravity alone, with a = −g = −9.81 m/s² (taking upward as positive). Air resistance is neglected in idealised free-fall problems.
Kinematic graphs
Plots of position vs. time, velocity vs. time, or acceleration vs. time. The relationships between these graphs (slopes, areas) are fundamental.
The three quantities form a chain linked by differentiation (going down) and integration (going up):
v(t) = dx/dt (velocity is the derivative of position)
a(t) = dv/dt = d²x/dt² (acceleration is the derivative of velocity)
x(t) = ∫v(t) dt (position is the integral of velocity)
v(t) = ∫a(t) dt (velocity is the integral of acceleration)
This means the slope of a position-time graph at any instant gives the instantaneous velocity, and the slope of a velocity-time graph gives the instantaneous acceleration.
Conversely, the area under a velocity-time curve over some interval gives the displacement during that interval, and the area under an acceleration-time curve gives the change in velocity.
Position vs. time (x-t):
Slope = velocity
Straight line = constant velocity
Curve (concave up) = positive acceleration
Curve (concave down) = negative acceleration (or positive deceleration)
Horizontal line = object at rest
Velocity vs. time (v-t):
Slope = acceleration
Area under curve = displacement
Line crossing zero = object changes direction
Horizontal line = constant velocity (zero acceleration)
Acceleration vs. time (a-t):
Area under curve = change in velocity
Horizontal line = constant acceleration
Zero line = constant velocity
These four equations apply only when acceleration is constant:
v = v₀ + at
x = x₀ + v₀t + ½at²
v² = v₀² + 2a(x − x₀)
x = x₀ + ½(v₀ + v)t
Each equation involves four of the five kinematic variables (x, v, v₀, a, t) and omits one. Choose the equation that includes the three knowns and the one unknown you need.
Strategy: list what you know and what you want, then pick the equation that connects them.
A projectile moves in two dimensions simultaneously. The key insight is that horizontal and vertical motions are independent:
Horizontal: a_x = 0, so v_x = v₀ cos θ (constant), x = x₀ + v_x · t
Vertical: a_y = −g, so standard kinematic equations apply with a = −g
Time links the two directions. Find time from whichever direction gives you enough information, then use it in the other direction.
When two objects move in the same reference frame, relative velocity is:
v_A relative to B = v_A − v_B
This matters in lab contexts where you measure motion relative to a sensor or camera rather than relative to the ground.
Kinematic equations (constant acceleration):
v = v₀ + at
x = x₀ + v₀t + ½at²
v² = v₀² + 2a(x − x₀)
x = x₀ + ½(v₀ + v)t
Free fall: a = −g = −9.81 m/s² (upward positive convention)
Projectile components: v₀x = v₀ cos θ , v₀y = v₀ sin θ
⚠️ Displacement and distance are different. A ball thrown up and caught at the same height has zero displacement but nonzero distance.
⚠️ Negative velocity does not mean "slowing down." It means moving in the negative direction. Slowing down means velocity and acceleration have opposite signs.
⚠️ The kinematic equations only work for constant acceleration. If acceleration changes with time, you need calculus (integration).
⚠️ In projectile motion, the horizontal velocity is constant (no horizontal acceleration if air resistance is neglected). At the peak of the trajectory, v_y = 0 but v_x is unchanged.
⚠️ Graph-reading questions are common. Practise extracting velocity from the slope of x-t graphs and displacement from the area under v-t graphs.
Q: A car accelerates from rest at 3 m/s² for 5 seconds. How far does it travel?
A: Using x = x₀ + v₀t + ½at²: x = 0 + 0 + ½(3)(5²) = 37.5 m.
Q: On a velocity-time graph, what does a straight line with negative slope represent?
A: Constant negative acceleration (the object is decelerating if it is moving in the positive direction, or speeding up if it is moving in the negative direction).
Q: A ball is thrown straight up at 20 m/s. How high does it go?
A: At the peak, v = 0. Using v² = v₀² + 2a(Δx): 0 = 20² + 2(−9.81)(Δx), so Δx = 400/19.62 ≈ 20.4 m.
Q: What is the area under a velocity-time curve?
A: The displacement of the object over that time interval.
Q: If an object's position-time graph is a parabola opening upward, what can you say about its acceleration?
A: The acceleration is constant and positive. A parabolic x-t curve corresponds to x = x₀ + v₀t + ½at² with a > 0.
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