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the vertical axis, while time is the horizontal variable used to show when each velocity occurs. The correct answer is Time."},{"text":"Acceleration.","correct":false,"explanation":"Acceleration is represented by the gradient of a velocity-time graph, not its horizontal axis. The horizontal axis shows Time."},{"text":"Displacement.","correct":false,"explanation":"Displacement is not the independent variable on this graph and would require a displacement-time graph. The horizontal axis represents Time."}]},{"id":"da2c032d-c27a-4586-9d1d-38ca80a81683","cardType":"basic","points":1,"front":"On a velocity-time graph, what is on the vertical axis?","back":"Velocity.","mcqOptions":[{"text":"Velocity.","correct":true,"explanation":"Velocity is plotted vertically so its value can be read at any particular time along the horizontal axis."},{"text":"Time.","correct":false,"explanation":"Time is the horizontal axis on a velocity-time graph. The vertical axis shows the measured velocity, so the correct answer is Velocity."},{"text":"Acceleration.","correct":false,"explanation":"Acceleration is calculated from the graph’s gradient rather than plotted directly on the vertical axis. The vertical axis represents Velocity."},{"text":"Displacement.","correct":false,"explanation":"Displacement is found from the area under a velocity-time graph, not from its vertical scale. The vertical axis shows Velocity."}]},{"id":"8692f3c4-3f9c-43c6-aad2-707705c0d0b3","cardType":"basic","points":1,"front":"If a journey starts from rest, what is the initial velocity?","back":"$0\\text{ m s}^{-1}$.","mcqOptions":[{"text":"$0\\text{ m s}^{-1}$.","correct":true,"explanation":"Starting from rest means the object initially has no motion, so its velocity at the start is exactly zero metres per second."},{"text":"$1\\text{ m s}^{-1}$.","correct":false,"explanation":"A non-zero value would mean the object was already moving. The phrase from rest means the initial velocity is $0\\text{ m s}^{-1}$."},{"text":"$9.8\\text{ m s}^{-1}$.","correct":false,"explanation":"The value $9.8\\text{ m s}^{-1}$ is often confused with gravitational acceleration, but an object starting from rest has initial velocity $0\\text{ m s}^{-1}$."},{"text":"$-0\\text{ m s}^{-1}$.","correct":false,"explanation":"Negative zero is not a distinct physical velocity, and introducing a negative sign is unnecessary. Starting from rest gives $0\\text{ m s}^{-1}$."}]},{"id":"494212e6-134d-44d9-ad6f-06695fb19775","cardType":"cloze","points":null,"front":"When velocity never goes below zero, its size is the [ ].","back":"When velocity never goes below zero, its size is the speed.","mcqOptions":[{"text":"speed","correct":true,"explanation":"When velocity remains non-negative, its magnitude equals its numerical value, so the size of the velocity is the speed."},{"text":"velocity","correct":false,"explanation":"Velocity includes direction, whereas the size or magnitude of velocity is speed. Therefore, the correct term for the blank is speed."},{"text":"displacement","correct":false,"explanation":"Displacement measures position change and is not the size of velocity. When velocity never becomes negative, its size is the speed."},{"text":"acceleration","correct":false,"explanation":"Acceleration describes how velocity changes, not the magnitude of velocity itself. The size of a non-negative velocity is called speed."}]},{"id":"4b21b523-faf7-419a-9439-8ef7acb71b46","cardType":"basic","points":1,"front":"What shape shows constant acceleration on a velocity-time graph?","back":"A straight line.","mcqOptions":[{"text":"A straight line.","correct":true,"explanation":"Constant acceleration means velocity changes by equal amounts in equal time intervals, producing a straight line with constant gradient."},{"text":"A horizontal line.","correct":false,"explanation":"A horizontal line has zero gradient, so it represents constant velocity and zero acceleration. Constant non-zero acceleration is shown by a straight line with a non-zero gradient."},{"text":"A curve.","correct":false,"explanation":"A curve has a changing gradient, indicating that acceleration is changing. Constant acceleration is represented by a straight line."},{"text":"A vertical line.","correct":false,"explanation":"A vertical line would represent an instantaneous change in velocity and does not describe ordinary constant acceleration. The graph should be a straight line."}]},{"id":"c45a8245-865c-4e91-87d2-76d3ae090251","cardType":"basic","points":1,"front":"What does a horizontal section mean for acceleration?","back":"Zero acceleration.","mcqOptions":[{"text":"Zero acceleration.","correct":true,"explanation":"A horizontal section has zero gradient, and acceleration is the gradient of a velocity-time graph. Therefore, the acceleration is zero."},{"text":"Constant acceleration.","correct":false,"explanation":"A horizontal section means velocity is constant, so its gradient is zero. The correct interpretation is Zero acceleration, not merely constant acceleration."},{"text":"Increasing acceleration.","correct":false,"explanation":"Increasing acceleration would require the gradient to become steeper over time. A horizontal section has zero gradient, giving Zero acceleration."},{"text":"Negative acceleration.","correct":false,"explanation":"Negative acceleration requires a negative gradient, shown by a falling section. A horizontal section has zero gradient and therefore Zero acceleration."}]},{"id":"27e224cc-4169-48b0-8282-13b4cbbe0703","cardType":"basic","points":1,"front":"On a velocity-time graph, what does a falling straight line represent if forward is positive?","back":"Negative acceleration.","mcqOptions":[{"text":"Negative acceleration.","correct":true,"explanation":"A falling line has a negative gradient. Since acceleration equals the gradient of a velocity-time graph, the acceleration is negative when forward is positive."},{"text":"Positive acceleration.","correct":false,"explanation":"Positive acceleration would produce a rising line because velocity increases with time. A falling line has a negative gradient, so it represents Negative acceleration."},{"text":"Zero acceleration.","correct":false,"explanation":"Zero acceleration is shown by a horizontal line with zero gradient. The falling line has a negative gradient, representing Negative acceleration."},{"text":"Changing acceleration.","correct":false,"explanation":"A straight line has a constant gradient, so its acceleration is constant rather than changing. Because the line falls, that constant acceleration is Negative acceleration."}]},{"id":"7f7c91a2-a992-43d2-aca3-730064ecd94a","cardType":"cloze","points":null,"front":"On a velocity-time graph, acceleration is found from the [ ].","back":"On a velocity-time graph, acceleration is found from the gradient.","mcqOptions":[{"text":"gradient","correct":true,"explanation":"Acceleration is the rate of change of velocity with time, which is the gradient of a velocity-time graph: change in velocity divided by change in time."},{"text":"intercept","correct":false,"explanation":"The intercept gives the velocity when time is zero, not the rate at which velocity changes. Acceleration is found from the gradient."},{"text":"area","correct":false,"explanation":"The area under a velocity-time graph gives displacement. Acceleration is determined by the graph’s gradient."},{"text":"y-coordinate","correct":false,"explanation":"The y-coordinate gives the velocity at a particular time, not its rate of change. Acceleration is calculated from the gradient."}]},{"id":"c7ee7f6d-413c-4609-b74d-22f1fbc2882a","cardType":"basic","points":1,"front":"What is the formula for acceleration from a velocity-time graph section?","back":"$a=\\frac{\\text{change in velocity}}{\\text{change in time}}$.","mcqOptions":[{"text":"$a=\\frac{\\text{change in velocity}}{\\text{change in time}}$.","correct":true,"explanation":"Acceleration is the rate of change of velocity, so divide the change in velocity by the corresponding change in time. The gradient of a velocity-time graph represents this quantity."},{"text":"$a=\\frac{\\text{change in time}}{\\text{change in velocity}}$.","correct":false,"explanation":"This reverses the gradient calculation, giving time change per velocity change rather than acceleration. Acceleration is found using change in velocity divided by change in time."},{"text":"$a=\\frac{\\text{velocity}}{\\text{time}}$.","correct":false,"explanation":"Using the whole velocity and time ignores the changes over the graph section. Acceleration depends on the change in velocity divided by the change in time."},{"text":"$a=\\frac{\\text{change in displacement}}{\\text{change in time}}$.","correct":false,"explanation":"This gives average velocity, because velocity is displacement divided by time. Acceleration instead requires the change in velocity divided by the change in time."}]},{"id":"06934100-cb37-4d27-bff8-6d169586872f","cardType":"basic","points":1,"front":"What are the units of acceleration?","back":"$\\text{m s}^{-2}$.","mcqOptions":[{"text":"$\\text{m s}^{-2}$.","correct":true,"explanation":"Acceleration measures how velocity changes each second. Since velocity has units of metres per second, dividing by another second gives metres per second squared, written as $\\text{m s}^{-2}$."},{"text":"$\\text{m s}^{-1}$.","correct":false,"explanation":"These are the units of velocity, not acceleration. Acceleration is the change in velocity per second, so its units include a second twice: $\\text{m s}^{-2}$."},{"text":"$\\text{m}^{2}\\text{ s}^{-1}$.","correct":false,"explanation":"Multiplying metres by metres would be appropriate for an area-related quantity, not acceleration. The correct units come from velocity divided by time: $\\text{m s}^{-2}$."},{"text":"$\\text{s}^{-2}\\text{ m}^{-1}$.","correct":false,"explanation":"This incorrectly puts metres in the denominator. Acceleration has metres in the numerator and seconds squared in the denominator, giving $\\text{m s}^{-2}$."}]},{"id":"40f5deb8-34b1-4570-9d87-30958becb2eb","cardType":"basic","points":1,"front":"The height of a velocity-time graph gives which quantity?","back":"Velocity, not acceleration.","mcqOptions":[{"text":"Velocity, not acceleration.","correct":true,"explanation":"The vertical coordinate on a velocity-time graph is velocity, while acceleration is represented by the gradient. Therefore, the graph’s height directly gives velocity."},{"text":"Acceleration, not velocity.","correct":false,"explanation":"This confuses height with gradient. Height gives the velocity value at that time, whereas acceleration is found from how steeply the graph rises or falls."},{"text":"Displacement, not velocity.","correct":false,"explanation":"Displacement is found from the signed area under the graph, not its vertical height. The height at a particular time gives the velocity."},{"text":"Distance, not velocity.","correct":false,"explanation":"Distance is obtained by adding absolute areas under a velocity-time graph. The vertical height itself represents velocity, including its possible positive or negative sign."}]},{"id":"ab06cbba-b4cd-4b21-893e-75e4e82e651d","cardType":"cloze","points":null,"front":"The [ ] under a velocity-time graph gives displacement.","back":"The area under a velocity-time graph gives displacement.","mcqOptions":[{"text":"area","correct":true,"explanation":"Displacement equals the signed area under a velocity-time graph. Areas above the time axis contribute positively, while areas below it contribute negatively."},{"text":"gradient","correct":false,"explanation":"The gradient of a velocity-time graph gives acceleration, not displacement. Displacement is calculated from the signed area between the graph and the time axis."},{"text":"height","correct":false,"explanation":"The graph’s height gives velocity at a particular time. Displacement requires accumulating velocity over time, which is represented by the area under the graph."},{"text":"intercept","correct":false,"explanation":"An intercept only shows a value where the graph crosses an axis and does not accumulate motion over time. Displacement is obtained from the signed area."}]},{"id":"ffad0784-00ee-430b-813d-e19ec162db56","cardType":"basic","points":1,"front":"When does distance equal displacement on a velocity-time graph?","back":"When there is no area below the time axis.","mcqOptions":[{"text":"When there is no area below the time axis.","correct":true,"explanation":"If the graph never goes below the time axis, velocity does not become negative. Every contribution to distance and displacement is then positive, so their values are equal."},{"text":"When there is no area above the time axis.","correct":false,"explanation":"No area above the axis would mean no positive displacement, but distance can still exceed displacement. They are equal when the graph has no area below the time axis."},{"text":"When there is equal area above and below the time axis.","correct":false,"explanation":"Equal positive and negative areas give zero displacement, while distance remains positive because distances use absolute areas. Equality requires no area below the time axis."},{"text":"When there is constant area below the time axis.","correct":false,"explanation":"Any area below the time axis represents motion in the opposite direction and makes distance larger than displacement. Equality requires no area below the time axis."}]},{"id":"7539d4b1-d4a2-4fda-bb03-f13a3b4ac4cd","cardType":"basic","points":1,"front":"For displacement, what does area below the time axis represent?","back":"Negative displacement.","mcqOptions":[{"text":"Negative displacement.","correct":true,"explanation":"Velocity below the time axis is negative, so its signed area contributes negatively to displacement. This represents displacement in the direction opposite to the chosen positive direction."},{"text":"Positive displacement.","correct":false,"explanation":"Area below the time axis corresponds to negative velocity, so it contributes negatively to displacement. Positive displacement comes from signed area above the time axis."},{"text":"Negative distance.","correct":false,"explanation":"Distance cannot be negative because it uses the total magnitude of motion. A below-axis region gives negative displacement, but its absolute area contributes positively to distance."},{"text":"Zero displacement.","correct":false,"explanation":"A below-axis region contributes negative displacement, but it only produces zero overall displacement if it exactly cancels all positive area elsewhere on the graph."}]},{"id":"a766a4bc-15ac-40fe-8c45-a8015f23c29f","cardType":"basic","points":2,"front":"How do you find total distance if part of a velocity-time graph is below the time axis?","back":"Add the absolute areas.","mcqOptions":[{"text":"Add the absolute areas.","correct":true,"explanation":"Distance measures total path length, so motion in either direction must contribute positively. Taking the absolute value of each region before adding prevents opposite motions from cancelling."},{"text":"Add the signed areas.","correct":false,"explanation":"Adding signed areas allows below-axis regions to cancel above-axis regions, which calculates displacement. Total distance requires treating every region as positive by adding absolute areas."},{"text":"Subtract the absolute areas.","correct":false,"explanation":"Subtracting absolute areas can make one section cancel another, producing neither total distance nor the required total magnitude of motion. Distance is found by adding all absolute areas."},{"text":"Add the gradient areas.","correct":false,"explanation":"The gradient gives acceleration and is not itself an area to add for distance. Distance comes from adding the absolute areas between the graph and the time axis."}]},{"id":"f4bed698-a4cc-4956-83c1-32485cc5b537","cardType":"basic","points":1,"front":"What units result from area under a velocity-time graph, using seconds and $\\text{m s}^{-1}$?","back":"Metres.","mcqOptions":[{"text":"Metres.","correct":true,"explanation":"Area under the graph has units of velocity multiplied by time: $(\\text{m s}^{-1})(\\text{s})=\\text{m}$. Therefore, it represents displacement in metres."},{"text":"Metres per second.","correct":false,"explanation":"Metres per second are the units of velocity, which is the graph’s vertical quantity. Multiplying velocity by time for an area leaves metres."},{"text":"Metres per second squared.","correct":false,"explanation":"Metres per second squared are the units of acceleration, found from the graph’s gradient. The area uses velocity multiplied by time and therefore has units of metres."},{"text":"Seconds.","correct":false,"explanation":"Seconds measure time along the horizontal axis, not the completed area. Combining seconds with velocity units, $\\text{m s}^{-1}\\times\\text{s}$, gives metres."}]},{"id":"fc2037e7-d296-41b5-8d2f-a82d047e9f89","cardType":"cloze","points":null,"front":"The area of a triangular velocity-time section is [ ].","back":"The area of a triangular velocity-time section is $\\frac{1}{2}\\times\\text{base}\\times\\text{height}$.","mcqOptions":[{"text":"$$\\frac{1}{2}\\times\\text{base}\\times\\text{height}$","correct":true,"explanation":"A triangular section has half the area of a rectangle with the same base and perpendicular height, so its area is calculated using $rac{1}{2}\\times\\text{base}\\times\\text{height}$."},{"text":"$$\\text{base}\\times\\text{height}$","correct":false,"explanation":"This uses the rectangle formula and therefore doubles the triangular area. The correct formula is $\\frac{1}{2}\\times\\text{base}\\times\\text{height}$ because a triangle occupies half the corresponding rectangle."},{"text":"$$\\frac{1}{2}\\times(\\text{base}+\\text{height})$","correct":false,"explanation":"Adding the dimensions does not calculate an area and gives incompatible units. A triangle’s area is $\\frac{1}{2}\\times\\text{base}\\times\\text{height}$, using the product of its perpendicular dimensions."},{"text":"$$\\frac{1}{4}\\times\\text{base}\\times\\text{height}$","correct":false,"explanation":"Halving both dimensions makes the result half the required triangular area. Only the product is halved, so the correct formula is $\\frac{1}{2}\\times\\text{base}\\times\\text{height}$."}]},{"id":"ba207807-c0eb-4691-a7ba-bfdbb31bc724","cardType":"cloze","points":null,"front":"A trapezium area is [ ], where $a$ and $b$ are parallel sides.","back":"A trapezium area is $\\frac{1}{2}(a+b)h$, where $a$ and $b$ are parallel sides.","mcqOptions":[{"text":"$$\\frac{1}{2}(a+b)h$","correct":true,"explanation":"A trapezium’s area equals its height multiplied by the mean length of its parallel sides. Their mean is $\\frac{a+b}{2}$, giving $\\frac{1}{2}(a+b)h$."},{"text":"$$(a+b)h$","correct":false,"explanation":"This omits the factor of one half and therefore doubles the trapezium’s area. The height must multiply the mean of the parallel sides, giving $\\frac{1}{2}(a+b)h$."},{"text":"$$\\frac{1}{2}(a-b)h$","correct":false,"explanation":"Subtracting the parallel sides does not find their mean and can even produce a negative result. Their lengths must be added and halved, giving $\\frac{1}{2}(a+b)h$."},{"text":"$$\\frac{1}{2}abh$","correct":false,"explanation":"Multiplying the parallel sides introduces an extra length factor and produces volume units rather than area units. The correct trapezium formula is $\\frac{1}{2}(a+b)h$."}]},{"id":"ce0ad593-997c-49c7-91ac-5b7267480c81","cardType":"basic","points":2,"front":"A tram goes from rest to $18\\text{ m s}^{-1}$ in $30\\text{ s}$. What is its acceleration?","back":"$0.6\\text{ m s}^{-2}$.","mcqOptions":[{"text":"$0.6\\text{ m s}^{-2}$.","correct":true,"explanation":"Acceleration is change in velocity divided by time. From rest, $a=\\frac{18-0}{30}=0.6\\text{ m s}^{-2}$, so this is the tram’s acceleration."},{"text":"$540\\text{ m s}^{-2}$.","correct":false,"explanation":"Multiplying velocity by time gives $540$, but acceleration requires division by time. The correct calculation is $18\\div30=0.6\\text{ m s}^{-2}$."},{"text":"$1.67\\text{ m s}^{-2}$.","correct":false,"explanation":"Dividing the time by the velocity reverses the acceleration formula. Acceleration is change in velocity divided by time, so $18\\div30=0.6\\text{ m s}^{-2}$."},{"text":"$6\\text{ m s}^{-2}$.","correct":false,"explanation":"This results from dividing $18$ by $3$ rather than by the full $30\\text{ s}$. Using $a=\\frac{\\Delta v}{\\Delta t}$ gives $0.6\\text{ m s}^{-2}$."}]},{"id":"bc6838a0-0e3a-446a-9308-ee08e6bc2e19","cardType":"basic","points":1,"front":"A car accelerates for $8\\text{ s}$, then drives at constant speed for $12\\text{ s}$. When does deceleration begin?","back":"$20\\text{ s}$.","mcqOptions":[{"text":"$20\\text{ s}$.","correct":true,"explanation":"The car spends $8\\text{ s}$ accelerating and then $12\\text{ s}$ at constant speed. Deceleration therefore begins at the cumulative time $8+12=20\\text{ s}$."},{"text":"$8\\text{ s}$.","correct":false,"explanation":"This marks the end of the acceleration phase, when constant-speed motion begins, not deceleration. Adding the following $12\\text{ s}$ gives a deceleration start time of $20\\text{ s}$."},{"text":"$12\\text{ s}$.","correct":false,"explanation":"This treats the duration of constant-speed travel as an absolute time measured from the start. It begins after $8\\text{ s}$, so deceleration starts at $8+12=20\\text{ s}$."},{"text":"$96\\text{ s}$.","correct":false,"explanation":"Multiplying the two phase durations does not locate an event on the timeline. Their durations must be added, so deceleration begins after $8+12=20\\text{ s}$."}]},{"id":"6ec4efa4-9a38-47bd-b327-d1ac7fc37470","cardType":"cloze","points":null,"front":"For the same change in velocity, a larger acceleration magnitude means a [ ] time.","back":"For the same change in velocity, a larger acceleration magnitude means a shorter time.","mcqOptions":[{"text":"shorter","correct":true,"explanation":"Since acceleration magnitude is $|a|=\\frac{|\\Delta v|}{\\Delta t}$, keeping the velocity change fixed means increasing acceleration requires the change to occur over a shorter time."},{"text":"longer","correct":false,"explanation":"This reverses the inverse relationship between acceleration and time. For a fixed velocity change, $\\Delta t=\\frac{|\\Delta v|}{|a|}$, so a larger acceleration magnitude means a shorter time."},{"text":"unchanged","correct":false,"explanation":"The time would remain unchanged only if the acceleration magnitude also remained unchanged for the same velocity change. A larger acceleration produces the fixed change in a shorter time."},{"text":"doubled","correct":false,"explanation":"Doubling the time would halve, rather than increase, the acceleration magnitude for a fixed velocity change. A larger acceleration therefore corresponds to a shorter time."}]},{"id":"c7ac7968-f4e4-4547-8078-f2a7bcf5e39b","cardType":"basic","points":2,"front":"A car accelerates from rest to $V$ in $8\\text{ s}$. Its deceleration magnitude is twice as large. How long does braking from $V$ to rest take?","back":"$4\\text{ s}$.","mcqOptions":[{"text":"$4\\text{ s}$.","correct":true,"explanation":"For the same velocity change, time is inversely proportional to acceleration magnitude. Doubling the magnitude therefore halves the original $8\\text{ s}$, giving a braking time of $4\\text{ s}$."},{"text":"$8\\text{ s}$.","correct":false,"explanation":"This assumes braking has the same acceleration magnitude as the initial acceleration. Since the deceleration magnitude is doubled, the same velocity change takes half of $8\\text{ s}$, or $4\\text{ s}$."},{"text":"$16\\text{ s}$.","correct":false,"explanation":"This incorrectly doubles the time when the acceleration magnitude doubles. For a fixed velocity change, time is inversely proportional to acceleration, so braking takes $8\\div2=4\\text{ s}$."},{"text":"$2\\text{ s}$.","correct":false,"explanation":"This divides the original time by four, as though the deceleration magnitude were four times larger. Doubling the magnitude only halves the time, giving $4\\text{ s}$."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"9.2 Velocity-Time Graphs","title":"Velocity-Time Graphs Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"d3b0951c-3945-4ac1-be22-0454dc7be718","walled":"$L39"}],"$L3a","$L3b"]

9.2 Velocity-Time Graphs Flashcards

9.2 Velocity-Time Graphs Flashcards