ards","topicId":"66efe5f0-1cf5-4a50-b137-d6158eb95d64","topicSlug":"kinematics-7","cards":[{"id":"cbd91743-ac6b-48dc-b84f-7efba6c6e366","cardType":"basic","front":"How is displacement found from initial and final position vectors?","back":"Displacement is $\\mathbf{s}=\\mathbf{r}-\\mathbf{r}_0$.","points":1,"mcqOptions":null},{"id":"b2677675-1681-4ff6-a575-3d6b245c82fd","cardType":"basic","front":"What scalar equations are represented by one two-dimensional vector equation?","back":"

One equation for the horizontal direction
One equation for the vertical direction
Both use the same time $t$

","points":3,"mcqOptions":null},{"id":"1e91600c-6305-40c3-befb-589ac5a0b007","cardType":"basic","front":"Why is average velocity $\\frac{1}{2}(\\mathbf{u}+\\mathbf{v})$ under constant acceleration?","back":"Velocity changes linearly from $\\mathbf{u}$ to $\\mathbf{v}$.","points":2,"mcqOptions":null},{"id":"011cb188-df47-4615-9db7-53679ec154ca","cardType":"cloze","front":"For constant acceleration, velocity is $\\text{[...]}$ and displacement is $\\mathbf{s}=\\mathbf{u}t+\\frac{1}{2}\\mathbf{a}t^2$.","back":"For constant acceleration, velocity is $\\boxed{\\mathbf{v}=\\mathbf{u}+\\mathbf{a}t}$ and displacement is $\\mathbf{s}=\\mathbf{u}t+\\frac{1}{2}\\mathbf{a}t^2$.","points":null,"mcqOptions":null},{"id":"44e945f3-2265-495a-87d1-61e8a1bea86a","cardType":"cloze","front":"For constant acceleration, velocity is $\\mathbf{v}=\\mathbf{u}+\\mathbf{a}t$ and displacement is $\\text{[...]}$.","back":"For constant acceleration, velocity is $\\mathbf{v}=\\mathbf{u}+\\mathbf{a}t$ and displacement is $\\boxed{\\mathbf{s}=\\mathbf{u}t+\\frac{1}{2}\\mathbf{a}t^2}$.","points":null,"mcqOptions":null},{"id":"b3cb449e-84b0-40ef-b1b5-e54e9bac6c84","cardType":"cloze","front":"With initial position $\\mathbf{r}_0$, the position after time $t$ is $\\text{[...]}$.","back":"With initial position $\\mathbf{r}_0$, the position after time $t$ is $\\boxed{\\mathbf{r}=\\mathbf{r}_0+\\mathbf{u}t+\\frac{1}{2}\\mathbf{a}t^2}$.","points":null,"mcqOptions":null},{"id":"7c508c56-1cad-44c0-80f2-784199c00a0b","cardType":"basic","front":"What does $v_y=0$ mean if $v_x\\neq0$?","back":"The particle is moving horizontally at that instant.","points":1,"mcqOptions":null},{"id":"3e159234-e382-4ed6-ac06-ebe91d7762ca","cardType":"basic","front":"For $\\mathbf{u}=\\begin{pmatrix}3\\\\8\\end{pmatrix}$, $\\mathbf{a}=\\begin{pmatrix}2\\\\-4\\end{pmatrix}$, find $\\mathbf{v}$ after $2$ s.","back":"$$\\mathbf{v}=\\begin{pmatrix}7\\\\0\\end{pmatrix}\\text{ m s}^{-1}$.","points":2,"mcqOptions":null},{"id":"1d2cb02e-f8a1-49e3-84d4-a3a434fa3817","cardType":"basic","front":"How must $v^2=u^2+2as$ be written in vector form?","back":"$$\\mathbf{v}\\mathbin{\\cdot}\\mathbf{v}=\\mathbf{u}\\mathbin{\\cdot}\\mathbf{u}+2\\mathbf{a}\\mathbin{\\cdot}\\mathbf{s}$.","points":1,"mcqOptions":null},{"id":"a191ca32-1181-4b86-9b6a-521c1ca7074a","cardType":"basic","front":"Why is $\\mathbf{v}^2=\\mathbf{u}^2+2\\mathbf{a}\\mathbf{s}$ invalid?","back":"The required ordinary vector multiplication is not defined.","points":2,"mcqOptions":null},{"id":"ae413535-7a65-417a-8af9-35bcfe5131a2","cardType":"basic","front":"Taking right and upward as positive, what is a projectile's acceleration?","back":"$$\\mathbf{a}=\\begin{pmatrix}0\\\\-g\\end{pmatrix}$, where $g=9.8\\text{ m s}^{-2}$.","points":1,"mcqOptions":null},{"id":"9f6b19c4-d7f8-4551-9f1e-93003516360f","cardType":"basic","front":"Why does an ideal projectile's horizontal velocity remain constant?","back":"Gravity provides no horizontal acceleration.","points":2,"mcqOptions":null},{"id":"429cca34-6bff-4832-a37e-352d9888bca7","cardType":"cloze","front":"For an ideal projectile, the displacement components are $\\text{[...]}$ and $y=u_yt-\\frac{1}{2}gt^2$.","back":"For an ideal projectile, the displacement components are $\\boxed{x=u_xt}$ and $y=u_yt-\\frac{1}{2}gt^2$.","points":null,"mcqOptions

4.8.1 Kinematics (A-level only) Flashcards

4.8.1 Kinematics (A-level only) Flashcards