Physics 2.8 - 2.10, 2.13 - Acceleration and velocity graphs

Acceleration

Speed describes how quickly distance is covered. Velocity includes a stated direction, so acceleration is defined using velocity rather than speed. Choose a positive direction before using signs; in straight-line examples, motion in the chosen direction has positive velocity.

The unit comes from $\frac{\mathrm{m\,s^{-1}}}{\mathrm{s}}=\mathrm{m\,s^{-2}}$. An acceleration of $2\,\mathrm{m\,s^{-2}}$ means the velocity changes by $2\,\mathrm{m\,s^{-1}}$ during each second while that acceleration remains uniform.

Worked example: positive acceleration

A cyclist's velocity increases from $2.0\,\mathrm{m\,s^{-1}}$ to $8.0\,\mathrm{m\,s^{-1}}$ in $3.0\,\mathrm{s}$.

1. Choose the forward direction as positive.
2. Identify $u = 2.0\,\mathrm{m\,s^{-1}}$, $v = 8.0\,\mathrm{m\,s^{-1}}$, $t = 3.0\,\mathrm{s}$.
3. Select $a = (v - u) / t$.
4. Substitute: $a = (8.0 - 2.0) / 3.0$.
5. Calculate: $a = 6.0 / 3.0 = 2.0\,\mathrm{m\,s^{-2}}$.
6. Sense-check: a change of $2.0\,\mathrm{m\,s^{-1}}$ in each of 3 seconds gives a total change of $6.0\,\mathrm{m\,s^{-1}}$, matching $8.0 - 2.0$.

If an object moving in the positive direction slows down, $v - u$ is negative. For example, changing from $18\,\mathrm{m\,s^{-1}}$ to $6\,\mathrm{m\,s^{-1}}$ in $4\,\mathrm{s}$ gives $(6 - 18) / 4 = -3\,\mathrm{m\,s^{-2}}$. The negative sign describes the direction of the acceleration relative to the chosen positive direction. It is often called deceleration in this situation.

An object can momentarily have zero velocity and still have non-zero acceleration. A ball at the highest point of an upward flight has zero velocity for an instant, but its free-fall acceleration is still downward.

Acceleration without time

This equation is required on both Foundation and Higher tiers.

When time is not given, uniform straight-line acceleration can connect initial velocity, final velocity and distance directly:

Use signs consistently. If forward is positive, braking while moving forward has negative acceleration. The squared velocities have units $(\mathrm{m\,s^{-1}})^2$, matching $2ax$, whose units are $\mathrm{m\,s^{-2}}\times\mathrm{m}=\mathrm{m^2\,s^{-2}}$.

Worked example: finding final velocity without time

A vehicle has initial velocity $4.0\,\mathrm{m\,s^{-1}}$ and accelerates uniformly at $3.0\,\mathrm{m\,s^{-2}}$ over $14\,\mathrm{m}$. Find its final velocity.

1. Identify $u = 4.0\,\mathrm{m\,s^{-1}}$, $a = 3.0\,\mathrm{m\,s^{-2}}$, $x = 14\,\mathrm{m}$.
2. Select $v^{2} - u^{2} = 2ax$.
3. Rearrange: $v^{2} = u^{2} + 2ax$.
4. Substitute: $v^{2} = 4.0^{2} + 2 \times 3.0 \times 14$.
5. Calculate: $v^{2} = 16 + 84 = 100\,\mathrm{m^2\,s^{-2}}$.
6. Take the square root: $v = 10\,\mathrm{m\,s^{-1}}$.
7. Choose the positive root because the vehicle continues in the defined forward direction. The final velocity is greater than the initial velocity, which is sensible for positive acceleration.

Use this equation for constant acceleration in a straight line. If the object reverses direction, distinguish signed displacement from total distance; the worked journey above does not reverse. Choose between the assigned relationships by asking whether time is known.

Velocity-time graphs

A velocity-time graph has time on the horizontal axis and velocity on the vertical axis. This changes the meaning of the same-looking line:

• a horizontal line above zero means constant non-zero velocity, so acceleration is zero;
• a rising straight line means uniform positive acceleration;
• a falling straight line means uniform negative acceleration;
• the steeper the line, the greater the magnitude of the acceleration.

The numerical gradient and area methods in this lesson use straight sections representing uniform acceleration.

For a graph that remains on or above the time axis, the area between the graph line and the time axis gives distance travelled. The area unit is $\mathrm{m\,s^{-1}}\times\mathrm{s}=\mathrm{m}$, confirming that the result is a distance.