15.4 The constant-acceleration formulae — Learn

Prerequisites: rearranging a formula; solving a linear equation; solving a quadratic equation; the gradient and area of a velocity–time graph from Section 15.3.

By the end of this section you can: list the five constant-acceleration quantities; derive the four formulae from a velocity–time graph; choose the formula that contains the quantities you know; use the formulae with a sign convention.

1. The five quantities

When an object moves in a straight line with constant acceleration, its motion is fixed by five quantities.

The word "initial" means at the start of the interval you are studying, and "final" means at the end of it. Each of $u$, $v$, $a$ and $s$ can be negative, because each is a vector along the line.

2. Where the formulae come from

The formulae all follow from the velocity–time graph. With constant acceleration, the graph is a straight line from $(0, u)$ to $(t, v)$.

The gradient of the line gives the acceleration:

Rearranging gives the first formula:

The area under the line is the displacement. The area is a trapezium with parallel sides $u$ and $v$ and height $t$, so

Substituting $v = u + at$ into that formula and simplifying gives

Eliminating $t$ between the first two formulae gives a fourth:

Rearranging $s = ut + \frac{1}{2}at^2$ for $ut$ and substituting gives a fifth form, $s = vt - \frac{1}{2}at^2$.

3. Choosing the formula

Each formula omits one of the five quantities. Find the quantity you do not know, and use the formula that omits it.

Example 1. A cyclist moves at $6$ m/s and accelerates at $3$ m/s$^2$ for $4$ s. Find the final velocity and the displacement.

The known quantities are $u = 6$, $a = 3$ and $t = 4$. Use $v = u + at$:

Now use $s = ut + \frac{1}{2}at^2$ with the same values:

Example 2. A car starts from rest and accelerates at $2.5$ m/s$^2$ over a distance of $45$ m. Find its final velocity and the time taken.

Here $u = 0$, $a = 2.5$ and $s = 45$. The time is not wanted yet, so use $v^2 = u^2 + 2as$:

Now use $v = u + at$ with $v = 15$:

Example 3. A train travelling at $30$ m/s brakes uniformly and stops in $90$ m. Find the acceleration and the time.

Here $u = 30$, $v = 0$ and $s = 90$. Use $v^2 = u^2 + 2as$:

Then use $v = u + at$:

4. A sign convention for vertical motion

For an object moving vertically, choose upwards or downwards as positive. Then give $u$, $v$, $a$ and $s$ their signs before you substitute. Gravity acts downwards, so if upwards is positive, $a = -9.8$ m/s$^2$.

Example 4. A ball is thrown vertically upwards at $14$ m/s. Find the greatest height it reaches. Take $g = 9.8$ m/s$^2$.

Take upwards as positive. At the greatest height the ball is instantaneously at rest, so $v = 0$. Use $v^2 = u^2 + 2as$ with $u = 14$ and $a = -9.8$:

The greatest height is $10$ m.

5. Try it: a ball thrown upwards

The two sliders set the launch speed $u$ and the time $t$ since the throw. The figure draws the ball at its height above the launch point, with the greatest height marked.

Predict first. Take $u = 14$ m/s and $a = -9.8$ m/s$^2$. Before you move the time slider, predict the greatest height the ball reaches and the velocity at that highest point. Write down both predictions.

Test. Move the time slider from $t = 0$ to the end of the flight and watch the ball, the velocity arrow and the height. Then set the launch speed to $20$ m/s and run the flight again.

Explain. Write two sentences. The first says why the velocity is zero at the greatest height while the acceleration stays at $-9.8$ m/s$^2$. The second says why the ball is at the same height at two different times.

Progress check

Return to the free kick. For the vertical motion, take upwards as positive. The ball leaves the ground with vertical velocity $u = 10\sqrt{2} \approx 14.1$ m/s, and the acceleration is $a = -9.8$ m/s$^2$. The ball reaches the goal line $t = 2\sqrt{2} \approx 2.83$ s after the kick.

1. Write down the formula that gives the height from $u$, $a$ and $t$.
2. Substitute the values and work out the height of the ball at the goal line.
3. Compare your answer with $2.44$ m, the height of the crossbar.
4. Use $v = u + at$ to find the vertical velocity of the ball at the goal line. Is the ball still rising or now falling?

Watch out. Three mistakes are common with these formulae.

The formula that contains the unknown. If the question asks for the time and the given values are $u$, $a$ and $s$, then $v^2 = u^2 + 2as$ cannot help, because it has no $t$ in it. List the quantities first, then pick the formula that omits the one you were not given. In Example 2 the first formula used was $v^2 = u^2 + 2as$, which omits $t$.

$u$ and $v$ swapped. $u$ is the velocity at the start and $v$ is the velocity at the end. In Example 3 the train starts at $30$ m/s, so $u = 30$ and $v = 0$. Swapping them makes the acceleration positive and the answer wrong.

A sign lost in a deceleration. With upwards or forwards positive, a deceleration is negative. In Example 3 the acceleration is $-5$ m/s$^2$, not $5$ m/s$^2$. Re-check: a positive $a$ with a positive $u$ must make $v$ larger, and a negative $a$ must make it smaller.

You should now be able to

• list the five constant-acceleration quantities and their units;
• derive $v = u + at$ and $s = \dfrac{u + v}{2}t$ from a velocity–time graph;
• choose the formula that omits the quantity you do not know;
• use the formulae with a sign convention;
• find a greatest height or a braking distance in a vertical or horizontal problem.