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Physics C: Mechanics Unit 1: Kinematics with Calculus

Kinematics describes how things move without asking why. In this course you use derivatives and integrals to move between position, velocity and acceleration, which every later unit depends on.

Covers: Position, velocity and acceleration as derivatives, Integrating to find velocity and position, Displacement versus distance, Constant-acceleration equations and free fall, Motion graphs, Projectile motion, Relative velocity

Study guide

Motion along a line is described by a position function x(t). Velocity is the rate of change of position, v = dx/dt, and acceleration is the rate of change of velocity, a = dv/dt = d²x/dt². Going the other way, you integrate: v(t) = v₀ + ∫a dt and x(t) = x₀ + ∫v dt. The constants of integration are the initial conditions, so always ask where the object started and how fast it was moving. Average velocity over an interval is Δx/Δt, while instantaneous velocity is the slope of the x–t graph at one moment. A speed is the magnitude of velocity and never negative.

Graphs carry the same information as the functions. The slope of an x–t graph is velocity, the slope of a v–t graph is acceleration, the area under a v–t graph is displacement, and the area under an a–t graph is the change in velocity. Area below the time axis counts as negative. That is why displacement and distance differ: displacement is ∫v dt, but distance is ∫|v| dt. To find distance, locate the times where v = 0, split the integral there, and add the absolute values of the pieces. An object speeds up when v and a have the same sign and slows down when they have opposite signs.

When acceleration is constant, integrating once and twice gives the familiar equations: v = v₀ + at, x = x₀ + v₀t + ½at², and v² = v₀² + 2aΔx. They are special cases, so use them only when a really is constant. Free fall near Earth's surface is the classic example, with a = −g ≈ −9.8 m/s² if up is positive. At the top of a vertical throw the velocity is zero but the acceleration is still g downward. If a depends on position instead of time, use the chain rule a = v dv/dx and separate variables.

In two dimensions, position becomes a vector r(t) = x(t) î + y(t) ĵ, and you differentiate each component separately. A projectile with no air resistance has aₓ = 0 and a_y = −g, so its horizontal velocity is constant while its vertical motion is free fall. The two directions are linked only by time. A standard strategy is to use the vertical motion to find the flight time and then multiply by vₓ. On level ground the range is R = v₀² sin(2θ)/g, maximized at 45°, and complementary angles give the same range. When the launch and landing heights differ, solve the full quadratic for time instead of using the range formula.

Velocity depends on the observer. If a boat moves at v_BW relative to the water and the water moves at v_WG relative to the ground, then v_BG = v_BW + v_WG as vectors. Draw the vector triangle, and decide which vector is the hypotenuse. For a plane that must travel in a particular direction across a wind, the airspeed is the hypotenuse and the ground velocity is a leg. Relative motion can also simplify problems: two objects that fall under gravity together have zero relative acceleration, so their separation changes at a constant rate.

Key ideas

• Differentiate position to get velocity and velocity to get acceleration; integrate to go back.
• Initial conditions supply the constants of integration.
• Displacement is ∫v dt, while distance is ∫|v| dt.
• Slope and area of motion graphs connect x, v and a.
• Constant-acceleration equations apply only when a is constant.
• Projectile motion splits into constant horizontal velocity and vertical free fall, linked by time.
• Relative velocities add as vectors.

Formulas and rules

Common mistakes

• Forgetting the initial position or initial velocity when integrating.
• Reporting net displacement when the question asks for total distance traveled.
• Using constant-acceleration equations when the acceleration depends on time or position.
• Thinking acceleration is zero at the top of a projectile's path.
• Using the level-ground range formula when the launch and landing heights differ.

Before the test, make sure you can

• Differentiate x(t) to find v(t) and a(t) at a given time.
• Integrate a(t) or v(t) with initial conditions to find velocity and position.
• Calculate total distance by splitting the integral where v = 0.
• Decide whether an object is speeding up or slowing down from the signs of v and a.
• Read slopes and areas from x–t, v–t and a–t graphs.
• Solve projectile problems on level ground and from a height.
• Use a = v dv/dx when velocity depends on position.
• Add velocities in different frames as vectors.

Words to know

Position

The location of an object relative to a chosen origin, x(t) in one dimension or r(t) in two or three.

Displacement

The change in position, Δx = x_final − x_initial; it can be negative.

Distance traveled

The total path length covered, ∫|v| dt; never negative.

Instantaneous velocity

The derivative dx/dt, the slope of the position–time graph at one moment.

Average velocity

Displacement divided by elapsed time, Δx/Δt.

Speed

The magnitude of velocity.

Acceleration

The rate of change of velocity, dv/dt = d²x/dt².

Initial conditions

The values x₀ and v₀ that fix the constants when integrating.

Free fall

Motion under gravity alone, with a = g ≈ 9.8 m/s² downward.

Projectile

An object moving under gravity alone after launch, with constant horizontal velocity.

Range

The horizontal distance a projectile travels before returning to its launch height.

Chain rule form of acceleration

a = v dv/dx, used when velocity is given as a function of position.

Relative velocity

The velocity of an object as measured by a particular observer; velocities between frames add as vectors.

Jerk

The rate of change of acceleration, da/dt.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.