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IMAT Physics · Kinematics

Equations of Motion, Motion Graphs and Free Fall

When the acceleration is constant, four equations link displacement, the two velocities, the acceleration and the time; graphs show the same motion as gradients and areas.

Why this matters

The 2019 paper asked for the top speed and the distance of a car that accelerates and then cruises. Constant-acceleration working also sits inside many dynamics and energy questions, so it is worth making automatic.

Concept 1 of 3: The equations of motion for constant acceleration

With constant acceleration the velocity rises by the same amount every second, so the average velocity is just the average of the start and end values. Distance is that average times the time. The other equations follow from these two ideas. Each equation leaves out one of the five quantities, so pick the one that skips the quantity you neither know nor want.

Definition

For motion in a straight line with constant acceleration (the five quantities are often called suvat):

  • v=u+atv = u + at (no ss)
  • s=12(u+v)ts = \tfrac{1}{2}(u + v)t (no aa)
  • s=ut+12at2s = ut + \tfrac{1}{2}at^2 (no vv)
  • v2=u2+2asv^2 = u^2 + 2as (no tt)

Choose a positive direction and give every vector quantity a sign. These equations do not apply when the acceleration changes; split the motion into stages instead.

Equations of uniformly accelerated motion

v=u+ats=ut+12at2v2=u2+2asv = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as
  • ssdisplacement, in m
  • u,vu, vinitial and final velocity, in m/s
  • aaconstant acceleration, in m/s²
  • tttime, in s

Worked example

A car moving at 4.0 m/s accelerates uniformly at 1.5 m/s21.5\ \text{m/s}^2 for 8.0 s. Find its final velocity and the distance covered.
Practice this conceptself-check · 4 quick reps

Stopping distance grows with the square of the speed

From v2=u2+2asv^2 = u^2 + 2as, the braking distance is u2/2au^2 / 2a. Double the speed and the distance is four times as long, not twice.

The equations hold only while the acceleration is constant

A car that accelerates and then cruises has two stages with different accelerations. Work out each stage on its own and carry the speed at the end of one into the next.

Concept 2 of 3: Motion graphs: gradients and areas

A graph is a picture of the whole motion. The steepness of a line tells you how fast the plotted quantity is changing, so the slope of a position graph is the velocity and the slope of a velocity graph is the acceleration. The area under a velocity graph adds up velocity × time strips, which is displacement.

Definition

On a displacement-time (s-t) graph:

  • the gradient is the velocity; a horizontal line means the body is at rest; a straight sloping line means constant velocity; a curve means the velocity is changing.

On a velocity-time (v-t) graph:

  • the gradient is the acceleration; a horizontal line means constant velocity; a straight sloping line means constant acceleration;
  • the area between the line and the time axis is the displacement (area below the axis counts as negative).

On an acceleration-time graph, the area is the change in velocity.

Reading a v-t graph

a=gradient=ΔvΔts=area under the linea = \text{gradient} = \frac{\Delta v}{\Delta t} \qquad s = \text{area under the line}
  • Δv/Δt\Delta v / \Delta trise over run of the v-t line

Worked example

A tram starts from rest and speeds up uniformly to 12 m/s in 4.0 s, travels at 12 m/s for 10 s, then slows uniformly to rest in 6.0 s. Sketch the v-t graph in your head and find the accelerations and the total distance.
Practice this conceptself-check · 3 quick reps

The same flat line means different things on different graphs

A horizontal line on an s-t graph means the body is standing still. On a v-t graph it means the body moves at a constant velocity. Read the axis labels before reading the shape.

Concept 3 of 3: Free fall and vertical throws

Near the Earth's surface, and with air resistance ignored, every object falls with the same constant acceleration, whatever its mass. So free fall is just constant acceleration with a=ga = g pointing down, and the equations of motion apply. A ball thrown straight up slows by about 10 m/s every second, stops for an instant, and comes back down as a mirror image of the way up.

Definition

  • In free fall the only force is gravity and the acceleration is g≈9.8 m/s2g \approx 9.8\ \text{m/s}^2 downwards (IMAT often says 10).
  • Without air resistance, all bodies fall with the same acceleration, whatever their mass.
  • For a vertical throw: at the highest point v=0v = 0 but a=ga = g downwards; the time up equals the time down; the ball returns at the speed it was thrown.
  • From rest: v=gtv = gt and h=12gt2h = \tfrac{1}{2}gt^2. Thrown up at uu: highest point h=u2/2gh = u^2 / 2g, reached after t=u/gt = u/g.

Falling from rest and the highest point of a throw

h=12gt2,v=gthmax⁡=u22gh = \tfrac{1}{2}gt^2, \quad v = gt \qquad h_{\max} = \frac{u^2}{2g}
  • ggacceleration of free fall, about 9.8 m/s²
  • hhheight fallen or risen, in m
  • uuspeed of an upward throw, in m/s

Worked example

A stone is dropped from rest from a bridge 45 m above a river. Taking g=10 m/s2g = 10\ \text{m/s}^2 and ignoring air resistance, how long does it fall and how fast does it hit the water?
Practice this conceptself-check · 4 quick reps

Heavier objects do not fall faster

Without air resistance every body falls with the same acceleration gg. A heavier body has a larger weight, but also more mass to accelerate, and the two cancel. Air resistance is what makes a feather fall slowly on Earth.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The equations of motion for constant acceleration

    Equations of uniformly accelerated motion

    v=u+ats=ut+12at2v2=u2+2asv = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as
  • Motion graphs: gradients and areas

    Reading a v-t graph

    a=gradient=ΔvΔts=area under the linea = \text{gradient} = \frac{\Delta v}{\Delta t} \qquad s = \text{area under the line}
  • Free fall and vertical throws

    Falling from rest and the highest point of a throw

    h=12gt2,v=gthmax⁡=u22gh = \tfrac{1}{2}gt^2, \quad v = gt \qquad h_{\max} = \frac{u^2}{2g}

Watch out for (4)

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