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displacement-time graph, a turning point has [ ], so the object is instantaneously stationary.","back":"On a displacement-time graph, a turning point has zero gradient, so the object is instantaneously stationary.","mcqOptions":null},{"id":"b63c41ff-0ebd-4a02-af5e-e7ed6a1a07cf","cardType":"cloze","points":null,"front":"On a displacement-time graph, a turning point has zero gradient, so the object is [ ].","back":"On a displacement-time graph, a turning point has zero gradient, so the object is instantaneously stationary.","mcqOptions":null},{"id":"78aed6c9-338d-4e71-abb5-43402dc00b01","cardType":"basic","points":1,"front":"What does crossing the time axis on a velocity-time graph usually indicate?","back":"A change of direction.","mcqOptions":null},{"id":"1b624d0b-e96a-4a67-a7e1-1f89873d1810","cardType":"basic","points":1,"front":"When does reaching $v=0$ not represent a change of direction?","back":"When velocity does not change sign.","mcqOptions":null},{"id":"88bc4039-d3f2-4f3f-af46-4233195a2b7b","cardType":"basic","points":2,"front":"How is distance travelled found from a velocity-time graph?","back":"Add the magnitudes of all areas, ignoring signs.","mcqOptions":null},{"id":"0ae58759-8230-401f-9992-0fb280a89bf1","cardType":"cloze","points":null,"front":"Areas above a velocity-time axis give [ ]; areas below it give negative displacement.","back":"Areas above a velocity-time axis give positive displacement; areas below it give negative displacement.","mcqOptions":null},{"id":"c10ff48c-993b-496e-aece-6df4529907a2","cardType":"cloze","points":null,"front":"Areas above a velocity-time axis give positive displacement; areas below it give [ ].","back":"Areas above a velocity-time axis give positive displacement; areas below it give negative displacement.","mcqOptions":null},{"id":"70dee28c-f683-4ed0-9d6f-01521e12f1e9","cardType":"basic","points":1,"front":"When do velocity and acceleration cause speed to increase?","back":"When they have the same sign.","mcqOptions":null},{"id":"5ead09fc-bdb9-443d-b4c0-d345353167d7","cardType":"basic","points":1,"front":"When do velocity and acceleration cause speed to decrease?","back":"When they have opposite signs.","mcqOptions":null},{"id":"d3b9583f-7c4c-40a9-9186-b51f59511850","cardType":"basic","points":2,"front":"Why can negative acceleration correspond to speeding up?","back":"If velocity is also negative, its magnitude increases.","mcqOptions":null},{"id":"a5168381-66b5-4dc4-b4ce-efedcc593fbf","cardType":"basic","points":1,"front":"How is average velocity found on a curved displacement-time graph?","back":"Find the gradient of the chord joining the endpoints.","mcqOptions":null},{"id":"3ff30397-94b6-4633-bb33-d28734f2a01b","cardType":"basic","points":1,"front":"How is instantaneous velocity found on a curved displacement-time graph?","back":"Find the gradient of the tangent at that point.","mcqOptions":null},{"id":"2573e6ac-9178-42aa-866c-2fa639fcf545","cardType":"cloze","points":null,"front":"Differentiating displacement gives $\\text{[...]}$, while differentiating velocity gives $a=\\frac{dv}{dt}$.","back":"Differentiating displacement gives $\\boxed{v=\\frac{ds}{dt}}$, while differentiating velocity gives $a=\\frac{dv}{dt}$.","mcqOptions":null},{"id":"6bc72032-a7da-472b-ac2b-f2e8c324c4ef","cardType":"cloze","points":null,"front":"Differentiating displacement gives $v=\\frac{ds}{dt}$, while differentiating velocity gives $\\text{[...]}$.","back":"Differentiating displacement gives $v=\\frac{ds}{dt}$, while differentiating velocity gives $\\boxed{a=\\frac{dv}{dt}}$.","mcqOptions":null},{"id":"4749d3e4-eb7c-4b9a-acb4-077c219aa58e","cardType":"basic","points":2,"front":"A particle moves from $s=5$ m at $t=2$ s to $s=23$ m at $t=8$ s along a straight graph section. 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