2.1 Scalars and vectors — Practise with help

1. $72\,\text{km}\,\text{h}^{-1} = 20\,\text{m}\,\text{s}^{-1}$.
2. $4.5 \times 10^{-4}\,\text{m}$.
3. The hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

If question 1 goes wrong, multiply by 1000 and divide by 3600. If question 2 goes wrong, count the places the decimal point moves: four places to the right after the 4. If question 3 goes wrong, use $c^2 = a^2 + b^2$, where $c$ is the hypotenuse.

Ask whether the quantity needs a direction to make sense in space. Energy and temperature do not; a force does.

Weight is a force, and acceleration has a direction. Energy and temperature have a magnitude only.

Energy and temperature are scalars. Acceleration and weight are vectors. Acceleration of $9.81\,\text{m}\,\text{s}^{-2}$ downwards is complete; a weight of 50 N needs a direction, such as 50 N downwards.

The two displacements are along the same line but in opposite directions. Choose north as positive.

Write 4.0 km as $+4.0$ and 2.5 km south as $-2.5$, then add the signed numbers.

The resultant displacement is 1.5 km north. The distance travelled is 6.5 km.

Look at what the two words mean. Speed is the magnitude of the velocity.

Say what happens to the direction of the car's motion as it goes round the bend.

The speed is a scalar, and it is the magnitude of the velocity. A steady speed of 15 m s⁻¹ means the magnitude does not change. The velocity is a vector, and its direction changes as the car goes round the bend. A change in direction is a change in velocity, so the velocity changes while the speed stays constant.

Take east as positive and give the westward displacement a negative sign.

Add $+8.0$ and $-6.0$.

The resultant displacement is 2.0 km east. The distance travelled is 14.0 km.

Think about how two currents combine at a junction, and compare that with the triangle rule for vectors.

State the test a quantity must pass to be a vector, then say whether current passes it.

A quantity is a vector only if it adds by the triangle rule. Two currents meeting at a junction add as ordinary numbers, and a current is not drawn as an arrow with a length. Current fails the vector test, so it is a scalar.

The two displacements are equal in size and opposite in direction.

Add $+50$ and $-50$.

The resultant displacement is zero, because the dog finishes where it started. The distance travelled is 100 m.