← Back to CoursesNewtonian Mechanics: Intermediate

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Kinematics in One Dimension

Manual: General · Subject: Physics

Analyze straight-line motion using position, velocity, and acceleration.

Describing straight-line motion

Position and displacement

In one dimension, motion can be described by a coordinate x(t)x(t). Displacement is Δx=xf−xi\Delta x = x_f - x_i. Positive and negative signs indicate direction relative to the chosen axis.

Velocity and acceleration

Average velocity is vavg=Δx/Δtv_{\text{avg}} = \Delta x/\Delta t. Instantaneous velocity is v=dx/dtv = dx/dt. Acceleration is a=dv/dt=d2x/dt2a = dv/dt = d^2x/dt^2. The sign of acceleration tells you how velocity changes, not necessarily whether the object speeds up or slows down.

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Common misconception

A negative acceleration does not always mean slowing down; if velocity is also negative, the object may be speeding up in the negative direction.

Constant-acceleration equations

When acceleration is constant, the motion equations are v=v0+atv = v_0 + at, x=x0+v0t+12at2x = x_0 + v_0 t + \tfrac{1}{2}at^2, and v2=v02+2a(x−x0)v^2 = v_0^2 + 2a(x-x_0). These are powerful because they connect kinematics without needing time in every problem.

Graphical interpretation

The slope of an xx-tt graph gives velocity. The slope of a vv-tt graph gives acceleration. The area under a vv-tt graph gives displacement, and the area under an aa-tt graph gives change in velocity.

If the slope of a velocity-time graph is constant and positive, what is happening?

Write the constant-acceleration equation that relates velocity and displacement without time.