Why can the suvat formulae only be used when the acceleration is constant?

The suvat formulae are derived from a velocity-time graph drawn as a straight line , and a straight line is exactly what constant acceleration gives. Every step of the derivation, the gradient and the areas alike, depends on that straight line. If the acceleration changes, the graph is a curve, the derivation no longer holds, and the motion has to be split into separate stages that each have constant acceleration.

True or False? A journey in which a car accelerates uniformly and then brakes to a stop can be handled by applying a single suvat formula to the journey as a whole.

False. The suvat formulae require the acceleration to be constant , and this journey has two different constant accelerations, so the formulae must be applied to each stage separately . The two stages are linked by the velocity between them: the final velocity of the accelerating stage is the initial velocity of the braking stage.

A constant-acceleration problem gives you only two of the five suvat quantities, so no single formula can be substituted into. What can you do instead?

Write down two of the suvat formulae and solve them as a pair of simultaneous equations . The five formulae are five different relations between the same five quantities, so any two of them containing your unknowns give two independent equations in those unknowns. Two unknowns need two equations, and a single formula can only ever supply one.

True or False? A ball thrown vertically upwards has zero speed at the moment it reaches the ground again.

False. The ball is still moving when it reaches the ground, and it is the impact that brings it to a stop, not the flight. For a ball moving freely under gravity, the only moment its speed is zero is at its highest point .

A LevelMathsEdexcelMechanicsFlashcardsKinematicsConstant Acceleration in 1D

Constant Acceleration in 1D (Edexcel A Level Maths: Mechanics): Flashcards

Exam code: 9MA0

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FrontDeriving the suvat Formulas
In the suvat formulae for motion in a straight line, what does each of the five letters stand for?
BackDeriving the suvat Formulas
The five quantities in the suvat formulae are:
$s$ , the displacement from the starting position
$u$ , the initial velocity
$v$ , the final velocity
$a$ , the acceleration
$t$ , the time taken
BackDeriving the suvat Formulas
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BackDeriving the suvat Formulas
Displacement, initial velocity, final velocity and acceleration are all vectors, so any of them can be negative, meaning the quantity points opposite to whichever direction has been chosen as positive.
So $u = - 6 {m s}^{- 1}$ describes an object that starts out moving in the negative direction at a speed of $6 {m s}^{- 1}$ .
Time is the one scalar, so a negative $t$ means an instant before the moment chosen as $t = 0$ , where the formulae still hold, letting you work backwards to before you started watching.

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Cards in this collection (21)

In the suvat formulae for motion in a straight line, what does each of the five letters stand for?
The five quantities in the suvat formulae are:
- $s$ , the displacement from the starting position
- $u$ , the initial velocity
- $v$ , the final velocity
- $a$ , the acceleration
- $t$ , the time taken
True or False?
In the suvat formulae, $s$ stands for the total distance the object has travelled.
False.
$s$ is the displacement from the starting position, not the distance travelled.
An object that moves away and comes back to where it started has $s = 0$ , however far it has actually travelled.
Which of the five suvat quantities can take a negative value, and what does a negative value mean?
Displacement, initial velocity, final velocity and acceleration are all vectors, so any of them can be negative, meaning the quantity points opposite to whichever direction has been chosen as positive.
So $u = - 6 {m s}^{- 1}$ describes an object that starts out moving in the negative direction at a speed of $6 {m s}^{- 1}$ .
Time is the one scalar, so a negative $t$ means an instant before the moment chosen as $t = 0$ , where the formulae still hold, letting you work backwards to before you started watching.
Why can the suvat formulae only be used when the acceleration is constant?
The suvat formulae are derived from a velocity-time graph drawn as a straight line, and a straight line is exactly what constant acceleration gives. Every step of the derivation, the gradient and the areas alike, depends on that straight line.
If the acceleration changes, the graph is a curve, the derivation no longer holds, and the motion has to be split into separate stages that each have constant acceleration.
On a velocity-time graph the velocity changes in a straight line from $u$ to $v$ over a time $t$ . Taking the area under the graph as a single shape, which shape is it, and which suvat formula does its area give?
The area under the graph is a trapezium with parallel sides of length $u$ and $v$ , a distance $t$ apart.
Its area gives:
$s = \frac{1}{2} (u + v) t$
which is the suvat formula connecting $s$ , $u$ , $v$ and $t$ .
If one of the two velocities is zero, the shape is a triangle rather than a trapezium, which is the same case with one parallel side of length zero, and the formula is unchanged.
The area under a velocity-time graph for constant acceleration can be split as a rectangle plus a triangle, or as a rectangle minus a triangle. Which suvat formula does each splitting give?
Both ways of splitting the area under the graph use a triangle of height $v - u$ , the change in velocity, on a base of $t$ .
A rectangle of height $u$ plus that triangle gives $s = u t + \frac{1}{2} (v - u) t$ , and a rectangle of height $v$ minus that triangle gives $s = v t - \frac{1}{2} (v - u) t$ .
Since $v - u = a t$ , these become:
$s = u t + \frac{1}{2} a t^{2}$
$s = v t - \frac{1}{2} a t^{2}$
The suvat formulae can also be derived using calculus. Fill in the two constants of integration:
$v = \int a d t = a t + c$
At $t = 0$ the velocity is $u$ , so $c =______$
$s = \int v d t = u t + \frac{1}{2} a t^{2} + c$
At $t = 0$ the displacement is $0$ , so $c =______$
Both constants of integration come from the values at $t = 0$ :
At $t = 0$ the velocity is $u$ , so $c = u$ , giving $v = u + a t$ .
At $t = 0$ the displacement is $0$ , so $c = 0$ , giving $s = u t + \frac{1}{2} a t^{2}$ .
The displacement is zero at the start because $s$ is measured from the starting position. Each constant is fixed by the state of the motion at the start, which is why $u$ appears in the first formula and nothing is added to the second.
Four of the five suvat formulae come straight from the velocity-time graph. Which one does not, and how is it obtained?
The formula that cannot be read off the velocity-time graph is:
$v^{2} = u^{2} + 2 a s$
It is the only one of the five that does not contain $t$ , so there is no time interval on the graph to obtain it from.
It is found instead by taking two of the other formulae and eliminating $t$ between them: rearranging $v = u + a t$ gives $t = \frac{v - u}{a}$ , and substituting that into $s = \frac{1}{2} (u + v) t$ leads to it.
In a constant-acceleration problem you know three of the five suvat quantities and want a fourth. How do you choose which of the five formulae to use?
Each of the five formulae contains four of the five quantities, so exactly one of them leaves out the quantity you neither know nor want.
Choose that one: the formula containing your three known values and the one you are looking for.
So with $u$ , $a$ and $t$ known and $s$ wanted, the quantity left out is $v$ , which points to $s = u t + \frac{1}{2} a t^{2}$ .
Constant-acceleration problems describe the values you need in words. Fill in the blanks with the suvat letter that each phrase pins down:
"… returns to its starting position …" means $______= 0$
"… initially at rest …" means $______= 0$
"… comes to rest …" means $______= 0$
The completed translations are:
"… returns to its starting position …" means $s = 0$ , because $s$ is measured from the starting position.
"… initially at rest …" means $u = 0$ .
"… comes to rest …" means $v = 0$ .
Before using the suvat formulae on a problem, why must you decide which direction counts as positive, and does it matter which direction you choose?
Until a positive direction is fixed, a value such as $u = - 6 {m s}^{- 1}$ has no meaning: the minus sign is what records the direction.
It does not matter which direction you choose, provided every quantity in the problem is measured against the same one. Choosing the direction the object starts out in, or the direction of the acceleration, usually leaves fewer negative values to handle.
True or False?
A journey in which a car accelerates uniformly and then brakes to a stop can be handled by applying a single suvat formula to the journey as a whole.
False.
The suvat formulae require the acceleration to be constant, and this journey has two different constant accelerations, so the formulae must be applied to each stage separately.
The two stages are linked by the velocity between them: the final velocity of the accelerating stage is the initial velocity of the braking stage.
A constant-acceleration problem gives you only two of the five suvat quantities, so no single formula can be substituted into. What can you do instead?
Write down two of the suvat formulae and solve them as a pair of simultaneous equations.
The five formulae are five different relations between the same five quantities, so any two of them containing your unknowns give two independent equations in those unknowns.
Two unknowns need two equations, and a single formula can only ever supply one.
A car speeding up along a straight road gives $v^{2} = 729$ from $v^{2} = u^{2} + 2 a s$ , so $v = \pm 27$ . How do you decide which sign to take?
Take the sign from the direction of travel, read against whichever direction was chosen as positive.
The car is speeding up in the direction it was already moving, so its final velocity has the same sign as its initial velocity, and taking that direction as positive gives $v = 27 {m s}^{- 1}$ .
Squaring has lost the direction, and only the situation being modelled can put it back: the negative root would describe an object moving the other way.
What value should you take for the acceleration due to gravity, and why might a question give you a different one?
Take $g = 9.8 {m s}^{- 2}$ unless a question specifies otherwise.
$g$ is not a universal constant: it depends on location, so it is not the same everywhere. A question is therefore free to specify a different value, $g = 10 {m s}^{- 2}$ being a common one, and a value given in the question always takes precedence.
A particle is moving vertically under gravity. Fill in the blanks with the acceleration in each case:
If you take upwards as the positive direction, $a =______$
If you take downwards as the positive direction, $a =______$
The completed statements are:
If you take upwards as the positive direction, $a = - g$
If you take downwards as the positive direction, $a = g$
Gravity always acts downwards, so the sign is decided by the direction you chose as positive, not by the way the particle happens to be moving. A ball on its way up still has $a = - g$ when upwards is positive.
A stone is dropped from the top of a cliff. What does that single word tell you about two of the suvat quantities?
Dropped means released from rest, so the initial velocity is $u = 0$ .
It also means the stone falls freely under gravity alone, so the acceleration is $g$ downwards, which is $a = 9.8 {m s}^{- 2}$ taking downwards as positive.
Thrown, projected and falling freely also signal that the acceleration is $g$ , but only dropped and released from rest give $u = 0$ as well.
A ball is thrown vertically upwards. What do you know about its velocity, and about its acceleration, at its highest point?
At the highest point the ball's velocity is instantaneously zero, which gives you $v = 0$ to use in the suvat formulae.
Its acceleration is unchanged: still $g$ downwards, as it is throughout the flight of a ball moving freely under gravity.
The ball does not stay at rest: its velocity is zero for that instant only, and the acceleration, which never stops acting, immediately starts it moving downwards.
True or False?
A ball thrown vertically upwards has zero speed at the moment it reaches the ground again.
False.
The ball is still moving when it reaches the ground, and it is the impact that brings it to a stop, not the flight.
For a ball moving freely under gravity, the only moment its speed is zero is at its highest point.
A ball is thrown vertically upwards from a window $h$ metres above the ground, and you want the time until it hits the ground. Taking upwards as positive, what value of $s$ do you use?
The displacement to use is $s = - h$ .
Displacement is measured from the starting position, which here is the window rather than the ground, and the ground is $h$ metres below it in the negative direction.
Had the ball been thrown from ground level and returned to the ground, the same reasoning would give $s = 0$ .
In a vertical-motion problem, what is gained by keeping $g$ as a symbol through the working instead of substituting $9.8$ at the start?
Keeping $g$ as a symbol often lets it cancel, so it never has to be evaluated at all.
Where it does not cancel, substituting only at the end keeps the working exact, so no accuracy is lost to rounding part-way through.
So, for example, a maximum height of $\frac{324}{2 g}$ metres is exact, and becomes $16.5$ metres, to 3 significant figures, only when $9.8$ is put in at the last step.

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