Changing velocity on a plane

Displacement and velocity for motion on a horizontal plane, finding change in velocity by vector subtraction when the direction changes, average acceleration in two dimensions, and why turning at constant speed is acceleration.

Drive round a roundabout at a steady 30 km/h and the speedometer needle never moves. You are still accelerating the whole way round, and your coffee knows it. Velocity has a direction, and changing direction changes velocity.

In a straight line, the change in velocity is just $v - u$ with signs. When the direction changes on a plane, $v - u$ has to be done as a vector subtraction. This page shows how, and how to get the average acceleration from it. It's also the starting point for circular motion in Year 12.

Change in velocity

$$\Delta\vec v = \vec v - \vec u = \vec v + (-\vec u)$$

Method for a vector diagram:

Draw $\vec v$.
From its head, draw $-\vec u$ (same length as $\vec u$, opposite direction).
$\Delta\vec v$ goes from the tail of $\vec v$ to the head of $-\vec u$.

Then the average acceleration is in the same direction as $\Delta\vec v$:

$$\vec a_{\text{avg}} = \frac{\Delta\vec v}{\Delta t}$$

A car turns left from north to west at a steady 15 m/s. The change in velocity points south-west, into the corner.

Worked example: turning a corner

A car travelling north at 15.0 m s⁻¹ turns a corner and is then travelling west at 15.0 m s⁻¹. The turn takes 4.0 s. Find the change in velocity and the average acceleration.

$\vec u = 15.0$ m s⁻¹ N, so $-\vec u = 15.0$ m s⁻¹ S. $\vec v = 15.0$ m s⁻¹ W.

vectors

$|\Delta\vec v| = \sqrt{15.0^2 + 15.0^2} = 21.2$ m s⁻¹.

v + (−u): perpendicular, so Pythagoras

West and south components are equal, so $\Delta\vec v$ points south-west (bearing 225°).

direction

$a = \dfrac{21.2}{4.0} = 5.30$ m s⁻² towards the south-west.

average acceleration

The speed didn't change at all, yet the car accelerated, towards the inside of the turn.

interpret

Worked example: a bounce

A tennis ball hits a wall at 8.0 m s⁻¹ east and rebounds at 6.0 m s⁻¹ west. It is in contact with the wall for 0.050 s. Find its average acceleration.

$u = +8.0$, $v = -6.0$.

east positive

$\Delta v = -6.0 - (+8.0) = -14.0$ m s⁻¹.

change in velocity

$a = \dfrac{-14.0}{0.050} = -280$ m s⁻², so 280 m s⁻² west. Note $\Delta v$ is 14, not 2: reversing direction is a big change.

acceleration

Using components instead

For awkward angles, find $\Delta v_x = v_x - u_x$ and $\Delta v_y = v_y - u_y$ separately, then recombine with Pythagoras and inverse tan. This is the same component method as 1F, applied to the subtraction.

Worked example: a yacht changing course

A yacht sailing at 6.0 m s⁻¹ on a bearing of 000° changes to 8.0 m s⁻¹ on a bearing of 060° over 20 s. Find its average acceleration.

$u_E = 0$, $u_N = 6.0$.

initial components

$v_E = 8.0\sin60° = 6.93$; $v_N = 8.0\cos60° = 4.00$.

final components

$\Delta v_E = 6.93$, $\Delta v_N = 4.00 - 6.0 = -2.00$.

differences

$|\Delta\vec v| = \sqrt{6.93^2 + 2.00^2} = 7.21$ m s⁻¹, at $\tan^{-1}(2.00/6.93) = 16°$ south of east: bearing 106°.

recombine

$a = 7.21 \div 20 = 0.36$ m s⁻² on a bearing of 106°.

acceleration

Displacement on a plane

Displacement on a plane follows the same rules. Add each leg of a journey as a vector, and average velocity is total displacement ÷ total time, in the direction of the displacement. For example, a car driving round three-quarters of a circular track of radius 50 m has travelled $\tfrac34 \times 2\pi(50) = 236$ m, but its displacement is a chord of length $\sqrt{50^2 + 50^2} = 70.7$ m.

InteractiveChange in velocity on a plane

Try this

Choose 90° left turn. Check: Δv is 21.2 m/s towards 225° and the average acceleration is 5.30 m/s², with no change in speed.
Choose U-turn (10 m/s north to 10 m/s south in 6 s). Why is the acceleration 3.33 m/s² south, when the speed is the same at the start and the end?
Choose Ball bouncing off a wall: 280 m/s² west. Increase the time to 0.10 s. What happens to the acceleration?
Use the sliders for the yacht example: 6 m/s at 000° to 8 m/s at 060° in 20 s. Check 7.2 m/s and 0.36 m/s².
Keep the speed fixed at 15 m/s and make the turn sharper and sharper. How does |Δv| change?

Test yourself

A car goes round a bend at a constant 20 m s⁻¹. Which statement is correct?
A puck moving at 4.0 m s⁻¹ east is hit and moves off at 3.0 m s⁻¹ north. Its change in velocity is
A ball hits a floor at 5.0 m s⁻¹ downwards and bounces up at 4.0 m s⁻¹. Contact lasts 0.020 s. The average acceleration during contact is
The average acceleration of an object always points in the direction of

Label the car turning the corner and its vector diagram.

Does the velocity change?

Change in velocity is found by vector : $\vec v - \vec u$. On a diagram, reverse $\vec u$ and add it to $\vec v$ head to . Average acceleration has the same direction as the change in . An object moving at constant speed around a bend is .

5 marks

A cyclist travelling at 6.0 m s⁻¹ east turns and travels at 8.0 m s⁻¹ south. The turn takes 2.5 s. Using a vector diagram, calculate the cyclist's average acceleration, and explain why it is not $(8.0 - 6.0)/2.5$.

Your answer