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

Constant Acceleration (Cambridge (CIE) A Level Maths: Mechanics): Flashcards

Exam code: 9709

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Frontsuvat in 1D
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?

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Cards in this collection (6)

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.

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