True or False? All five of the suvat quantities are vector quantities.

False. Displacement, initial velocity, final velocity and acceleration are all vectors , but time is a scalar . Because the other four are vectors, each of them carries a sign that records its direction.

Why do the suvat formulae only work when the acceleration is constant?

Because they are all derived from a velocity-time graph that is a straight line , and only a constant acceleration produces a straight line. If the acceleration varies the graph curves, and neither its gradient nor the region beneath it can be handled by the simple methods the derivations rely on.

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 instant it hits the ground.

False. The ball is still moving when it reaches the ground; it is the impact that brings it to rest, and the impact is not part of the motion being modelled. For an object moving freely under gravity, the speed is zero only at the highest point .

A ball is thrown vertically upwards and caught again at the height it was thrown from, so its displacement is zero. How far has it actually travelled?

It has travelled twice the maximum height : that height going up, and the same height again coming back down. Distance counts every metre the ball moves, whichever way it is going, which is why it can be large when the displacement is zero.

Why can the constant-acceleration formulae be used for an object moving freely under gravity?

Because gravity gives the object the same acceleration throughout its motion, so the acceleration is constant , which is exactly what those formulae require. Any other force that would change it, chiefly air resistance, is modelled as negligible , and that is what the word freely signals.

ASMathsCambridge (CIE)MechanicsFlashcardsKinematics (Straight Line Motion)Constant Acceleration

Constant Acceleration (Cambridge (CIE) AS Maths: Mechanics): Flashcards

Exam code: 9709

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FrontDeriving the suvat Formulae
Fill in the two missing letters from the suvat list:
$s$ is the displacement, $______$ is the initial velocity, $______$ is the final velocity, $a$ is the acceleration and $t$ is the time.

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Cards in this collection (22)

Fill in the two missing letters from the suvat list:
$s$ is the displacement, $______$ is the initial velocity, $______$ is the final velocity, $a$ is the acceleration and $t$ is the time.
$s$ is the displacement, $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration and $t$ is the time.
$u$ comes before $v$ in the alphabet, which is a reliable way of remembering which one is the velocity at the start.
True or False?
All five of the suvat quantities are vector quantities.
False.
Displacement, initial velocity, final velocity and acceleration are all vectors, but time is a scalar.
Because the other four are vectors, each of them carries a sign that records its direction.
On a velocity-time graph for constant acceleration the line runs from $(0, u)$ to $(t, v)$ , and its gradient is the acceleration. How does that give $v = u + a t$ ?
The gradient of the line is $\frac{v - u}{t}$ , so $a = \frac{v - u}{t}$ .
Multiplying both sides by $t$ gives $a t = v - u$ , which rearranges to $v = u + a t$ .
The area between a velocity-time graph and the time axis is the displacement. For constant acceleration, what shape is that area and what formula does it give?
It is a trapezium whose parallel sides are $u$ and $v$ and whose width is $t$ , so its area gives
$s = \frac{1}{2} (u + v) t$
If $u$ or $v$ is zero the shape is a triangle, which is this same formula with one parallel side of length zero.
The same region under the graph can be split into a rectangle and a triangle instead. What formula does that give?
The rectangle has area $u t$ and the triangle has base $t$ and height $v - u$ , so together they give $s = u t + \frac{1}{2} (v - u) t$ .
Since $v - u = a t$ , that becomes $s = u t + \frac{1}{2} a t^{2}$ .
Splitting the same region a third way, as a rectangle minus a triangle, gives $s = v t - \frac{1}{2} a t^{2}$ .
Which suvat formula is found by eliminating $t$ , and how?
$v^{2} = u^{2} + 2 a s$ , the only one of the five with no $t$ in it.
Rearranging $v = u + a t$ gives $t = \frac{v - u}{a}$ , and substituting that into $s = \frac{1}{2} (u + v) t$ gives $2 a s = (u + v) (v - u)$ .
Expanding the right-hand side gives $2 a s = v^{2} - u^{2}$ , which rearranges to the result.
Integrating a constant acceleration $a$ with respect to time gives $v = a t + c$ . What is $c$ , and why?
$c$ is $u$ , the initial velocity, so the result is $v = u + a t$ .
Putting $t = 0$ into $v = a t + c$ leaves $v = c$ , and the velocity when $t = 0$ is what $u$ means.
True or False?
The suvat formulae can be used for an object that is slowing down.
True.
A deceleration is simply a negative acceleration once the direction of motion is taken as positive, and nothing in the formulae requires $a$ to be positive.
They work with signed quantities throughout, so a negative value of $a$ is handled exactly like a positive one.
Why do the suvat formulae only work when the acceleration is constant?
Because they are all derived from a velocity-time graph that is a straight line, and only a constant acceleration produces a straight line.
If the acceleration varies the graph curves, and neither its gradient nor the region beneath it can be handled by the simple methods the derivations rely on.
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.
Define acceleration due to gravity.
The acceleration due to gravity, written $g$ , is the acceleration of an object moving freely under gravity alone, and it always acts vertically downwards.
Its true value varies slightly with location, so it is modelled by a fixed number: take $g = 10 {m s}^{- 2}$ unless you are told otherwise.
An object is moving freely under gravity. Fill in the two accelerations:
If upwards is taken as the positive direction, then $a =______$
If downwards is taken as the positive direction, then $a =______$
If upwards is taken as the positive direction, then $a = - g$ .
If downwards is taken as the positive direction, then $a = g$ .
Gravity always acts downwards, so the sign is decided entirely by which direction you chose to call positive.
A toy rocket is projected vertically upwards from the ground at $18 {m s}^{- 1}$ , and at its highest point its velocity is zero. Taking upwards as positive and $g = 10 {m s}^{- 2}$ , find the maximum height.
The maximum height is $16.2 m$ .
Here $u = 18 {m s}^{- 1}$ , $v = 0 {m s}^{- 1}$ and $a = - 10 {m s}^{- 2}$ , so $v^{2} = u^{2} + 2 a s$ gives $0 = 324 - 20 s$ .
Solving that gives $s = 16.2 m$ .
True or False?
A ball thrown vertically upwards has zero speed at the instant it hits the ground.
False.
The ball is still moving when it reaches the ground; it is the impact that brings it to rest, and the impact is not part of the motion being modelled.
For an object moving freely under gravity, the speed is zero only at the highest point.
A rocket projected upwards from the ground returns to the ground, giving $0 = t (36 - 10 t)$ , so $t = 0 s$ or $t = 3.6 s$ . Why is $t = 0 s$ rejected?
$t = 0 s$ is the instant the rocket was projected, when it was also at ground level.
The equation is satisfied at both moments the rocket is on the ground, so the root that answers the question is $t = 3.6 s$ .
A ball is thrown vertically upwards and caught again at the height it was thrown from, so its displacement is zero. How far has it actually travelled?
It has travelled twice the maximum height: that height going up, and the same height again coming back down.
Distance counts every metre the ball moves, whichever way it is going, which is why it can be large when the displacement is zero.
Why can the constant-acceleration formulae be used for an object moving freely under gravity?
Because gravity gives the object the same acceleration throughout its motion, so the acceleration is constant, which is exactly what those formulae require.
Any other force that would change it, chiefly air resistance, is modelled as negligible, and that is what the word freely signals.

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