Velocity Calculator

All four answer a different question, and mixing them up is the usual reason a homework answer disagrees with the key.

Average velocity — displacement over the whole interval:

$v_{\text{avg}} = \frac{\Delta x}{\Delta t}$

Instantaneous velocity — the value at a single instant, the derivative of position:

$v(t) = \frac{dx}{dt}$

Final velocity — the velocity at the end of an interval of constant acceleration:

$v = v_0 + at$

Change in velocity — a difference, not a rate:

$\Delta v = v_f - v_i$

Units: $x$ in metres (m), $t$ in seconds (s), $v$ in metres per second (m/s), $a$ in m/s².

The assumption people forget: $v = v_0 + at$ needs constant acceleration. Average and instantaneous velocity carry no such condition — they are definitions.

Position-time graph: velocity is the slope. The slope of the straight line joining two points is the average velocity between them; the slope of the tangent at a point is the instantaneous velocity there. A horizontal line means the object is at rest.

Velocity-time graph: the slope is acceleration ($a = \Delta v/\Delta t$, in m/s²) and the area under the curve is displacement, in metres. Area below the time axis counts as negative displacement.

$\Delta x = \int_{t_1}^{t_2} v(t)\,dt$

When it applies: the area rule is exact for any $v(t)$, curved or straight — it is just integration.

The assumption people forget: the area gives displacement, not distance. To get distance travelled you must add the absolute values of the areas above and below the axis separately.

• Averaging speeds instead of using total displacement — for a two-leg trip, $v_{\text{avg}}$ is total displacement over total time, not the mean of the two leg velocities, unless the legs take equal times.
• Reporting distance-based speed as velocity — an out-and-back trip has non-zero average speed and zero average velocity.
• Forgetting the sign of $\Delta v$ — slowing down from $25$ m/s to $10$ m/s gives $\Delta v = -15$ m/s. The minus sign is the deceleration.
• Using a tangent slope as an average — the tangent gives the value at one instant only.
• Unit drift — km/h and m/s differ by a factor of $3.6$; convert before, not after, the arithmetic.
• Reading a negative acceleration as slowing down — it speeds up an object that is already travelling in the negative direction. Compare the signs of velocity and acceleration: same sign means speeding up, opposite signs mean slowing down.
• Expecting the units to disambiguate — speed and velocity are both quoted in metres per second, so only the wording of the question tells you which one is wanted.