SQA Higher Physics

Distance, displacement, speed and velocity

A scalar has size only; a vector has size and direction. Distance and speed are scalars. Displacement and velocity are vectors, and so is acceleration.

The distance travelled is the total length of the path, however it twists. The displacement is the straight-line distance from the start to the finish, in a stated direction. Walk 30 m east and then 10 m west and the distance is 40 m but the displacement is 20 m east.

Speed and velocity are found in the same way, with the distance or displacement divided by the time taken:

Here $d$ is the distance in metres (m), $s$ is the displacement in metres, $v$ is the speed or velocity in metres per second ($\mathrm{m\,s^{-1}}$) and $t$ is the time in seconds (s). Used over a whole trip these give an average. The speed at one instant is found from a very short time interval.

Acceleration is the rate of change of velocity, in metres per second squared ($\mathrm{m\,s^{-2}}$). It is a vector, so an object that slows down in the direction it is moving has an acceleration in the opposite direction, and an object that changes direction at constant speed is also accelerating.

Motion with constant acceleration

For an object moving in a straight line with constant acceleration, five relationships link the quantities involved.

$u$ is the initial velocity, $v$ the final velocity, $a$ the constant acceleration, $t$ the time and $s$ the displacement, all in SI units. The fifth relationship is $s = vt$ on its own, which holds when $v$ is constant and $a = 0$.

The first is the definition of acceleration rearranged, since $a = (v - u)/t$. The fourth says that the displacement is the average velocity times the time, which is true when the velocity changes at a steady rate. Substituting $v = u + at$ into the fourth gives the second, and eliminating $t$ between the first and fourth gives the third.

Each of the four contains four of the five quantities $s, u, v, a, t$. So the method is to write down the three quantities the question gives, find which one is missing and unwanted, and choose the equation that leaves it out. If $a$ is unknown and not wanted, use $s = \tfrac{1}{2}(u+v)t$. If $t$ is neither given nor wanted, use $v^2 = u^2 + 2as$.

Signs

Every quantity in these equations is a vector, so each needs a sign. Choose one direction as positive, and give any quantity pointing the other way a negative sign. For an object thrown upwards it is usual to take up as positive, so the acceleration due to gravity is $a = -9.8\,\mathrm{m\,s^{-2}}$ throughout the flight, both on the way up and on the way down.

The equations give displacement, not distance. If the motion changes direction, the displacement at the end is smaller than the distance travelled, and the distance has to be found by splitting the motion at the point where the velocity is zero.

Worked examples

A car accelerating

A car speeds up from $4.0\,\mathrm{m\,s^{-1}}$ to $16\,\mathrm{m\,s^{-1}}$ over a distance of 60 m in a straight line. The quantities given are $u = 4.0$, $v = 16$ and $s = 60$, and the acceleration and time are wanted. Time is not given, so use $v^2 = u^2 + 2as$:

The time follows from $v = u + at$, so $t = (16 - 4.0)/2.0 = 6.0\,\mathrm{s}$. As a check, $s = \tfrac{1}{2}(4.0 + 16) \times 6.0 = 60\,\mathrm{m}$, which agrees with the data.

A ball thrown upwards

A ball is thrown vertically upwards at $14\,\mathrm{m\,s^{-1}}$. Take up as positive, so $u = +14$ and $a = -9.8$. At the highest point the velocity is momentarily zero, so $v = 0$.

The time to reach the top is $t = (v - u)/a = (0 - 14)/(-9.8) = 1.4\,\mathrm{s}$ (1.43 s before rounding). The ball returns to its starting height after twice that, because the motion down is the motion up in reverse, so the total flight time is 2.9 s. On return its velocity is $-14\,\mathrm{m\,s^{-1}}$, the same speed in the opposite direction.

Motion that changes direction

An object has an initial velocity of $+8.0\,\mathrm{m\,s^{-1}}$ and a constant acceleration of $-2.0\,\mathrm{m\,s^{-2}}$. After 7.0 s its velocity is

so it is moving back towards its start. Its displacement after 7.0 s is

which is 7.0 m from the start in the original direction. The distance travelled is larger. The object stops when $t = 8.0/2.0 = 4.0\,\mathrm{s}$, having travelled $\tfrac{1}{2}(8.0)(4.0) = 16\,\mathrm{m}$ forwards. In the remaining 3.0 s it accelerates from rest in the opposite direction, covering $\tfrac{1}{2}(2.0)(3.0)^2 = 9.0\,\mathrm{m}$. The distance is $16 + 9.0 = 25\,\mathrm{m}$, and the displacement is $16 - 9.0 = 7.0\,\mathrm{m}$, agreeing with the equation.

Choosing and checking

• Units must be consistent: metres, seconds and metres per second. A speed in kilometres per hour is converted first, by dividing by 3.6.
• Gravitational acceleration near the Earth's surface is taken as $9.8\,\mathrm{m\,s^{-2}}$ downwards, for objects where air resistance can be ignored.
• If the answer to an equation is a square root, the negative root is meaningful where direction matters. A ball thrown up passes a given height once going up and once coming down, and the two velocities there have the same size and opposite signs.
• A vertical motion and a horizontal motion can each be handled with these equations separately, as they are for a projectile.