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Physics HL · Chapter 1: Kinematics

1.2 Uniformly Accelerated Motion and Kinematics Equations

Develop instantaneous rates, then use constant-acceleration equations with correct sign handling.

Estimated time: 32 minutes

Instantaneous Velocity from the Position Graph

When motion is non-uniform, a single average velocity is not enough. We use instantaneous velocity: the rate of change of position at one moment. Geometrically, it is the slope of the tangent to the position-time curve at that point.

As you move along a curved position-time graph, tangent slope can change from positive to zero to negative. That changing slope is exactly why we need acceleration.

Acceleration as the Rate of Change of Velocity

a=ΔvΔt,v=u+ata = \frac{\Delta v}{\Delta t},\qquad v=u+at

For constant acceleration, the velocity-time graph is a straight line and its slope is constant.

Positive acceleration means velocity increases in the positive direction. Negative acceleration means velocity shifts toward the negative direction. Depending on current motion direction, negative acceleration may correspond to speeding up or slowing down, so always reason with signs, not keywords.

Displacement from Area Under a Velocity-Time Graph

For any interval, the signed area under a velocity-time graph equals displacement. Under constant acceleration, that area is a trapezoid, which gives a fast path to displacement relations and explains why average velocity in that interval is (u+v)/2.

Δs=u+v2t,Δs=ut+12at2,v2=u2+2aΔs\Delta s=\frac{u+v}{2}t,\qquad \Delta s=ut+\frac{1}{2}at^2,\qquad v^2=u^2+2a\Delta s

Use the equation that matches your known values instead of forcing one formula every time.

These equations are valid only when acceleration is constant over the time interval used. If acceleration changes with time, switch to graph methods or split the motion into intervals where acceleration can be treated as constant.

Simulation: Constant-Acceleration Equation Check

Set u, a, and t and compare computed outputs to graph-based displacement and velocity.

Uniform Acceleration Lab

v(t)

6.80 m/s

s(t)

10.80 m

v(final)

16.60 m/s

s(final)

92.70 m

x-t Curve

Test Yourself

A body has u = 4.0 m/s and v = 12.0 m/s after t = 6.0 s under constant acceleration. Enter the displacement in meters.

Hint: Use the average-velocity form with constant acceleration.