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Constant Acceleration in 2D (AQA A Level Maths: Mechanics): Flashcards

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Frontsuvat in 2D
Four of the five suvat formulae can be written with vectors. Which one cannot, and how do you use it in two dimensions instead?

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Four of the five suvat formulae can be written with vectors. Which one cannot, and how do you use it in two dimensions instead?
The one that cannot is $v^{2} = u^{2} + 2 a s$ .
It squares the velocities, and a vector cannot be squared, so there is no vector version of it.
Apply it to each component separately instead: one equation using only the $i$ components, and one using only the $j$ components.
Fill in the two missing words about the quantities in the vector suvat formulae:
$s$ , $u$ , $v$ and $a$ are all $______$ , but $t$ is a $______$ .
$s$ , $u$ , $v$ and $a$ are all vectors, but $t$ is a scalar.
That is why the four vector formulae multiply a vector by $t$ rather than combining two vectors, and why $t$ is never written in bold.
A question says a particle is travelling parallel to $(3 i - j)$ . What does that tell you about its velocity?
Its velocity is a scalar multiple of that vector, so it can be written $v = k (3 i - j)$ for some number $k$ .
That introduces one unknown, but it fixes the ratio of the two components, which is usually what makes the problem solvable.
True or False?
A particle moving parallel to $i$ has a velocity whose $j$ component is zero.
True.
Moving parallel to $i$ means moving purely horizontally, so the velocity has no vertical part at all.
It works the same way round the other way: a particle moving parallel to $j$ has an $i$ component of zero.
In a two-dimensional problem the acceleration $a$ is constant. What does that require beyond its magnitude staying the same?
Its direction must stay the same as well.
A vector is constant only when its magnitude and its direction are both unchanged, so a particle that turns while keeping its speed does not have constant acceleration, and the suvat formulae cannot be used on it.
A drone starts from rest and accelerates uniformly, reaching a velocity of $(0.6 i - 1.8 j) {m s}^{- 1}$ after 60 seconds. Find its acceleration.
The acceleration is $(0.01 i - 0.03 j) {m s}^{- 2}$ .
Starting from rest means $u = 0$ , so $v = u + a t$ gives $a = \frac{v}{t}$ .
Dividing each component by 60 gives $0.6 \div 60 = 0.01$ and $- 1.8 \div 60 = - 0.03$ .
What are the two ways of setting up the working in a two-dimensional suvat problem?
Either keep one vector equation throughout, collecting the $i$ and $j$ terms as you go, or split it into two separate equations, one for each component, and solve them independently.
Both routes give the same answer, so the choice is a matter of which is tidier for the question in front of you.

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