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It is calculated as change in velocity divided by the time taken."},{"text":"Acceleration is the rate of change of distance.","correct":false,"explanation":"Distance describes how far an object travels, whereas acceleration concerns changing velocity. The rate of change of distance is speed, not acceleration."},{"text":"Acceleration is the rate of change of speed.","correct":false,"explanation":"Speed changing can indicate acceleration, but acceleration is defined using velocity, which also includes direction. The complete definition is the rate of change of velocity."},{"text":"Acceleration is the rate of change of force.","correct":false,"explanation":"Force may cause acceleration, but acceleration does not measure how force changes. It measures the change in velocity per unit time."}]},{"id":"eb2c49eb-34bd-4387-b98b-3699b134cdf5","cardType":"basic","points":1,"front":"What equation calculates average acceleration?","back":"$$$a = \\dfrac{\\Delta v}{t}$$","mcqOptions":[{"text":"$$$a = \\dfrac{\\Delta v}{t}$$","correct":true,"explanation":"Average acceleration equals the change in velocity, $\\Delta v$, divided by the time taken, $t$. The units are metres per second squared."},{"text":"$$$a = \\dfrac{t}{\\Delta v}$$","correct":false,"explanation":"This reverses the required division and gives time per velocity rather than velocity change per time. Acceleration is calculated using $a = \\dfrac{\\Delta v}{t}$."},{"text":"$$$a = \\Delta v t$$","correct":false,"explanation":"Multiplying by time gives the wrong relationship and incorrect units. Average acceleration is the change in velocity divided by time: $a = \\dfrac{\\Delta v}{t}$."},{"text":"$$$a = \\dfrac{v}{t^2}$$","correct":false,"explanation":"This uses velocity rather than change in velocity and introduces an unnecessary square on time. The correct equation is $a = \\dfrac{\\Delta v}{t}$."}]},{"id":"2805ec2b-ef5f-41d8-9366-301fb337b98a","cardType":"basic","points":1,"front":"What is the unit of acceleration?","back":"m/s²","mcqOptions":[{"text":"m/s²","correct":true,"explanation":"Acceleration is change in velocity divided by time. Since velocity is measured in m/s, dividing by seconds gives m/s²."},{"text":"m/s","correct":false,"explanation":"m/s is the unit of velocity, not acceleration. Acceleration includes an additional division by time, giving m/s²."},{"text":"m²/s","correct":false,"explanation":"The squared unit belongs to distance here rather than the time interval. Acceleration has metres per second squared, written m/s²."},{"text":"N/kg","correct":false,"explanation":"N/kg is equivalent to m/s² in contexts involving force and mass, but the requested standard unit for acceleration here is m/s²."}]},{"id":"9e701759-4eae-4862-9267-76d0c3b94a14","cardType":"basic","points":1,"front":"How is change in velocity calculated?","back":"Final velocity - initial velocity","mcqOptions":[{"text":"Final velocity - initial velocity","correct":true,"explanation":"Change in velocity is found by subtracting the initial velocity from the final velocity. Keeping this order preserves the correct sign for acceleration."},{"text":"Initial velocity - final velocity","correct":false,"explanation":"This reverses the subtraction and changes the sign of the change in velocity. The correct calculation is final velocity minus initial velocity."},{"text":"Final velocity + initial velocity","correct":false,"explanation":"Adding the velocities does not measure the difference between them and can give an incorrect result. Change in velocity requires final minus initial velocity."},{"text":"Final velocity ÷ initial velocity","correct":false,"explanation":"Dividing velocities produces a ratio rather than a change in velocity. The change is calculated by subtracting initial velocity from final velocity."}]},{"id":"8ad7f7e2-01eb-4c70-ab93-75798cf81ccc","cardType":"basic","points":1,"front":"What does an acceleration of 3 m/s² mean?","back":"The velocity changes by 3 m/s every second.","mcqOptions":[{"text":"The velocity changes by 3 m/s every second.","correct":true,"explanation":"An acceleration of 3 m/s² means velocity changes by 3 m/s during each second. The change can be positive or negative depending on the acceleration's direction."},{"text":"The velocity changes by 3 m/s every 3 seconds.","correct":false,"explanation":"This treats the numerical acceleration as though it applies over three seconds. The denominator in m/s² means the velocity changes by 3 m/s each second."},{"text":"The velocity changes by 3 m every second.","correct":false,"explanation":"This gives a change in distance rather than velocity and has units of speed. Acceleration of 3 m/s² changes velocity by 3 m/s each second."},{"text":"The velocity changes by 9 m/s every second.","correct":false,"explanation":"Multiplying 3 m/s² by the example value of three seconds would calculate a total change over three seconds, not the change per second."}]},{"id":"285e5fc2-b189-48f4-9d57-289fec3e948b","cardType":"basic","points":2,"front":"A cyclist accelerates from 2 m/s to 8 m/s in 3 s. What is the average acceleration?","back":"$$$a = \\dfrac{8 - 2}{3} = 2\\ \\text{m/s}^2$$","mcqOptions":[{"text":"$$$a = \\dfrac{8 - 2}{3} = 2\\ \\text{m/s}^2$$","correct":true,"explanation":"Use $a = \\dfrac{v-u}{t}$. Substituting 8 m/s, 2 m/s and 3 s gives $\\dfrac{6}{3} = 2\\ \\text{m/s}^2$."},{"text":"$$$a = \\dfrac{8 + 2}{3} = \\dfrac{10}{3}\\ \\text{m/s}^2$$","correct":false,"explanation":"Adding the velocities instead of finding their change gives an incorrect numerator. Acceleration uses final velocity minus initial velocity, so the answer is 2 m/s²."},{"text":"$$$a = \\dfrac{8 - 2}{2} = 3\\ \\text{m/s}^2$$","correct":false,"explanation":"Using the initial velocity as the time accidentally replaces 3 s with 2 s. Dividing the 6 m/s change by 3 s gives 2 m/s²."},{"text":"$$$a = \\dfrac{8}{3} = 2.67\\ \\text{m/s}^2$$","correct":false,"explanation":"This omits the initial velocity, treating the final velocity as the total change. The velocity change is 8 - 2 = 6 m/s, giving 2 m/s²."}]},{"id":"090ba473-4149-41c8-aa87-ee29279cd9d6","cardType":"basic","points":2,"front":"Why does an object slowing down in the positive direction have negative acceleration?","back":"The acceleration is negative because the change in velocity is negative.","mcqOptions":[{"text":"The acceleration is negative because the change in velocity is negative.","correct":true,"explanation":"While velocity remains positive, slowing down makes final velocity smaller than initial velocity. Therefore final minus initial velocity is negative, giving negative acceleration."},{"text":"The acceleration is positive because the velocity is still positive.","correct":false,"explanation":"The sign of acceleration depends on the change in velocity, not simply whether velocity is positive. Slowing down in the positive direction produces a negative change."},{"text":"The acceleration is zero because the direction is unchanged.","correct":false,"explanation":"Unchanged direction does not mean unchanged velocity. Since the speed is decreasing, velocity changes and acceleration is non-zero and negative."},{"text":"The acceleration is negative because the velocity is negative.","correct":false,"explanation":"The acceleration is negative, but the object can still have positive velocity while slowing down. The negative sign comes from the negative change in velocity."}]},{"id":"88c2232a-e7bc-4f10-b2d1-6344158c8d69","cardType":"basic","points":1,"front":"A car has an acceleration of -5 m/s² while slowing down. What is the magnitude of its deceleration?","back":"The magnitude is the positive size of the deceleration: 5 m/s².","mcqOptions":[{"text":"The magnitude is the positive size of the deceleration: 5 m/s².","correct":true,"explanation":"Magnitude means size without direction or sign. The acceleration is -5 m/s², so its deceleration magnitude is the positive value 5 m/s²."},{"text":"The magnitude is the signed deceleration: -5 m/s².","correct":false,"explanation":"-5 m/s² is the signed acceleration, not its magnitude. Magnitude removes the negative sign, so the deceleration's size is 5 m/s²."},{"text":"The magnitude is the positive size of the deceleration: 25 m/s².","correct":false,"explanation":"Squaring the acceleration is an unnecessary operation and changes the units. Magnitude only removes the sign, leaving 5 m/s²."},{"text":"The magnitude is the positive size of the deceleration: 0.2 m/s².","correct":false,"explanation":"Taking the reciprocal gives a different quantity and reverses the numerical relationship. The magnitude of -5 m/s² is simply 5 m/s²."}]},{"id":"c24e6be5-a373-4a9c-ace0-16dc18fc7e45","cardType":"basic","points":1,"front":"How is acceleration found from a velocity-time graph?","back":"The gradient of the velocity-time graph.","mcqOptions":[{"text":"The gradient of the velocity-time graph.","correct":true,"explanation":"Gradient means change in vertical velocity divided by change in horizontal time. This is exactly change in velocity per unit time, which is acceleration."},{"text":"The area under the velocity-time graph.","correct":false,"explanation":"The area under a velocity-time graph gives displacement, not acceleration. Acceleration is found from the graph's gradient."},{"text":"The gradient of the distance-time graph.","correct":false,"explanation":"The gradient of a distance-time graph gives speed. Acceleration is obtained from the gradient of a velocity-time graph."},{"text":"The intercept of the velocity-time graph.","correct":false,"explanation":"An intercept gives a velocity value at a particular time, often the initial velocity. Acceleration depends on the graph's gradient."}]},{"id":"09941f29-5546-4e22-ae92-a28510e9c4c3","cardType":"basic","points":2,"front":"What does a horizontal line on a velocity-time graph show?","back":"A horizontal line shows constant velocity, so acceleration is zero.","mcqOptions":[{"text":"A horizontal line shows constant velocity, so acceleration is zero.","correct":true,"explanation":"A horizontal line has zero gradient. Since acceleration is the gradient of a velocity-time graph, zero gradient means constant velocity and zero acceleration."},{"text":"A horizontal line shows constant acceleration, so velocity is zero.","correct":false,"explanation":"A horizontal velocity-time line represents constant velocity, not necessarily zero velocity. Its zero gradient means acceleration is zero."},{"text":"A horizontal line shows increasing velocity, so acceleration is positive.","correct":false,"explanation":"Increasing velocity requires an upward slope on a velocity-time graph. A horizontal line has no change in velocity, so acceleration is zero."},{"text":"A horizontal line shows decreasing velocity, so acceleration is negative.","correct":false,"explanation":"Decreasing velocity requires a downward slope. A horizontal line keeps velocity constant, giving zero rather than negative acceleration."}]},{"id":"87ff8bc5-4a45-49c2-8a84-12bd8cf97a20","cardType":"basic","points":1,"front":"What does a curved section on a velocity-time graph indicate?","back":"A curved line shows changing acceleration.","mcqOptions":[{"text":"A curved line shows changing acceleration.","correct":true,"explanation":"The gradient of a velocity-time graph represents acceleration. A curved line has a changing gradient, so the acceleration changes with time."},{"text":"A curved line shows constant acceleration.","correct":false,"explanation":"Constant acceleration requires a constant gradient, which produces a straight line. A changing gradient on a curve indicates changing acceleration."},{"text":"A curved line shows constant velocity.","correct":false,"explanation":"Constant velocity is represented by a horizontal line. A curve means the velocity is changing with a non-constant gradient."},{"text":"A curved line shows zero acceleration.","correct":false,"explanation":"Zero acceleration requires zero gradient and therefore a horizontal line. A curved section has changing gradients, so its acceleration is generally changing."}]},{"id":"73f39a9e-78da-43eb-8ed4-222a1bc7b6fe","cardType":"basic","points":1,"front":"What does the area under a velocity-time graph represent?","back":"The area between the line and the time axis.","mcqOptions":[{"text":"The area between the line and the time axis.","correct":true,"explanation":"The area under a velocity-time graph is found between the plotted line and the time axis. It represents displacement, with regions below the axis carrying a negative sign."},{"text":"The gradient between the line and the time axis.","correct":false,"explanation":"Gradient measures acceleration on a velocity-time graph, while the enclosed area represents displacement. The two graph properties must not be confused."},{"text":"The height between the line and the time axis.","correct":false,"explanation":"Height gives a velocity difference at a particular time, not a distance-related quantity. The area between the line and time axis represents displacement."},{"text":"The length between the line and the time axis.","correct":false,"explanation":"A length alone does not combine velocity with time and therefore cannot represent displacement. Displacement is calculated from the graph's area."}]},{"id":"36dd7b46-456c-4f00-9461-4edb8a8d46cb","cardType":"basic","points":2,"front":"How is total distance found when part of a velocity-time graph is below the time axis?","back":"Add the magnitudes of all areas, including areas below the time axis.","mcqOptions":[{"text":"Add the magnitudes of all areas, including areas below the time axis.","correct":true,"explanation":"Total distance counts every part of the journey as a positive amount. Therefore use the magnitudes of all areas, including regions below the time axis."},{"text":"Add the signed values of all areas, including areas below the time axis.","correct":false,"explanation":"Adding signed areas gives displacement because areas below the axis are negative. Total distance instead requires their magnitudes to be added positively."},{"text":"Add the magnitudes of areas above the time axis, excluding areas below it.","correct":false,"explanation":"This ignores motion in the opposite direction. Areas below the axis still represent distance travelled, so their magnitudes must also be included."},{"text":"Subtract the magnitudes of all areas, including areas below the time axis.","correct":false,"explanation":"Subtracting areas can cancel parts of the journey and produce displacement-like results. Distance counts the whole journey, so all area magnitudes are added."}]},{"id":"e55397aa-90da-4568-9286-535f4374467d","cardType":"basic","points":1,"front":"What graph shape represents uniform acceleration on a velocity-time graph?","back":"A straight line with a constant gradient.","mcqOptions":[{"text":"A straight line with a constant gradient.","correct":true,"explanation":"Uniform acceleration means velocity changes by equal amounts in equal times. On a velocity-time graph this produces a straight line with constant gradient."},{"text":"A curved line with a constant gradient.","correct":false,"explanation":"A curve normally has a changing gradient, so it represents changing acceleration. Constant gradient on this graph is shown by a straight line."},{"text":"A horizontal line with a zero gradient.","correct":false,"explanation":"A horizontal line represents zero acceleration, which is a special case of constant velocity rather than the general graph shape for uniform acceleration."},{"text":"A straight line with a changing gradient.","correct":false,"explanation":"A straight line has the same gradient throughout. A changing gradient indicates changing acceleration and would require a curved graph."}]},{"id":"24064da5-15aa-49e8-b3f2-898a3a9b7cd5","cardType":"basic","points":2,"front":"When should $v^2-u^2=2as$ be used?","back":"Use it when acceleration is uniform and three of $u$, $v$, $a$ and $s$ are known.","mcqOptions":[{"text":"Use it when acceleration is uniform and three of $u$, $v$, $a$ and $s$ are known.","correct":true,"explanation":"The equation $v^2-u^2=2as$ is a constant-acceleration equation. It is useful when acceleration is uniform and three of the four variables are known."},{"text":"Use it when acceleration is changing and three of $u$, $v$, $a$ and $s$ are known.","correct":false,"explanation":"The equation assumes constant acceleration, so using it when acceleration changes can give an invalid result. Uniform acceleration is required."},{"text":"Use it when acceleration is uniform and two of $u$, $v$, $a$ and $s$ are known.","correct":false,"explanation":"Knowing only two variables is insufficient to determine the unknowns in this relationship. The stated condition requires three known variables."},{"text":"Use it when acceleration is uniform and three of $u$, $v$, $a$ and $t$ are known.","correct":false,"explanation":"This equation contains displacement $s$, not time $t$. The relevant known quantities are three of $u$, $v$, $a$ and $s$."}]},{"id":"4d5a432a-5010-4e30-8b1c-b495857adf26","cardType":"basic","points":1,"front":"Which equation links initial velocity, final velocity, acceleration and distance for uniform acceleration?","back":"$$$v^2-u^2=2as$$","mcqOptions":[{"text":"$$$v^2-u^2=2as$$","correct":true,"explanation":"This constant-acceleration equation links initial velocity $u$, final velocity $v$, acceleration $a$, and displacement $s$ without using time."},{"text":"$$$v-u=2as$$","correct":false,"explanation":"This omits the squares on the velocities and gives inconsistent units. The correct relationship is $v^2-u^2=2as$."},{"text":"$$$v^2+u^2=2as$$","correct":false,"explanation":"Adding the squared velocities is a sign error. The equation uses their difference, because initial velocity must be subtracted from final velocity squared."},{"text":"$$$v^2-u^2=2at$$","correct":false,"explanation":"This incorrectly replaces distance $s$ with time $t$. The equation linking these velocities, acceleration and distance uses $2as$."}]},{"id":"1fb1d788-a65f-4245-b178-745bd9dcc630","cardType":"basic","points":1,"front":"What is the acceleration of a freely falling object near Earth's surface?","back":"Approximately 9.8 m/s² downwards.","mcqOptions":[{"text":"Approximately 9.8 m/s² downwards.","correct":true,"explanation":"Near Earth’s surface, gravitational field strength is approximately 9.8 N/kg, so a freely falling object accelerates at about 9.8 m/s² downwards when air resistance is negligible."},{"text":"Approximately 9.8 m/s² upwards.","correct":false,"explanation":"The magnitude is correct, but the direction is wrong. Gravity pulls freely falling objects downwards, so their acceleration is approximately 9.8 m/s² downwards near Earth’s surface."},{"text":"Approximately 4.9 m/s² downwards.","correct":false,"explanation":"4.9 m/s² is half of the gravitational acceleration and can result from incorrectly dividing by two. The correct acceleration near Earth’s surface is approximately 9.8 m/s² downwards."},{"text":"Approximately 98 m/s² downwards.","correct":false,"explanation":"This is ten times the accepted value, usually caused by a powers-of-ten or unit conversion error. The gravitational acceleration near Earth’s surface is approximately 9.8 m/s² downwards."}]},{"id":"43fae951-e3a2-412a-8688-9fe621faefcd","cardType":"basic","points":3,"front":"What two main forces act on an object falling through a fluid?","back":"Weight acts downwards and drag acts upwards.","mcqOptions":[{"text":"Weight acts downwards and drag acts upwards.","correct":true,"explanation":"A falling object in a fluid experiences weight due to gravity downwards and drag from the fluid opposing its motion upwards."},{"text":"Weight acts upwards and drag acts downwards.","correct":false,"explanation":"The directions have been reversed. Weight always acts downwards near Earth, while drag opposes the object’s downward motion and therefore acts upwards."},{"text":"Thrust acts downwards and drag acts upwards.","correct":false,"explanation":"Thrust is a driving force produced by engines or propulsion, not the force causing an ordinary object to fall. The relevant downward force is weight, with drag upwards."},{"text":"Weight acts downwards and lift acts upwards.","correct":false,"explanation":"Lift is associated mainly with aerodynamic wings and is not the standard opposing force for an object falling through a fluid. The upward force is drag."}]},{"id":"e796ec68-375f-4c91-9dc5-2dbd780610b7","cardType":"basic","points":3,"front":"What happens to the forces and acceleration when terminal velocity is reached?","back":"At terminal velocity, drag equals weight, so the resultant force is zero and acceleration is zero.","mcqOptions":[{"text":"At terminal velocity, drag equals weight, so the resultant force is zero and acceleration is zero.","correct":true,"explanation":"When drag has increased to equal the object’s weight, the opposing forces balance. The resultant force and acceleration are then zero, so the object continues at constant velocity."},{"text":"At terminal velocity, drag is greater than weight, so the resultant force is upwards and acceleration is zero.","correct":false,"explanation":"If drag were greater than weight, there would be an upward resultant force and the object would accelerate upwards. Terminal velocity occurs when drag equals weight, making acceleration zero."},{"text":"At terminal velocity, drag is less than weight, so the resultant force is downwards and acceleration is zero.","correct":false,"explanation":"A smaller drag force leaves a downward resultant force, so the object would still accelerate. At terminal velocity, drag equals weight and the resultant force is zero."},{"text":"At terminal velocity, drag equals weight, so the resultant force is zero and acceleration is maximum.","correct":false,"explanation":"Balanced forces produce zero resultant force, and Newton’s second law therefore gives zero acceleration. Maximum acceleration occurs earlier, before drag has grown to equal weight."}]},{"id":"94d41249-024c-4759-912a-ff85cf3fcfd9","cardType":"cloze","points":null,"front":"Acceleration is measured in $\\text{[...]}$.","back":"Acceleration is measured in $\\boxed{\\text{m/s}^2}$.","mcqOptions":[{"text":"$$\\text{m/s}^2$","correct":true,"explanation":"Acceleration is the change in velocity per unit time, so its SI unit is metres per second per second, written as m/s²."},{"text":"$$\\text{m/s}$","correct":false,"explanation":"m/s is the unit of velocity, not acceleration. Acceleration measures how quickly velocity changes, so an additional division by time gives the unit m/s²."},{"text":"$$\\text{N}$","correct":false,"explanation":"The newton is the unit of force. Acceleration is measured from the change in velocity over time and has the unit m/s²."},{"text":"$$\\text{m}^2/\\text{s}$","correct":false,"explanation":"m²/s represents area divided by time, not change in velocity divided by time. The correct acceleration unit is m/s²."}]},{"id":"d7e1fd6a-8164-4bee-b659-b6fb7a98a57f","cardType":"cloze","points":null,"front":"On a velocity-time graph, the [ ] represents acceleration.","back":"On a velocity-time graph, the gradient represents acceleration.","mcqOptions":[{"text":"gradient","correct":true,"explanation":"The gradient of a velocity-time graph is change in velocity divided by time. This is the definition of acceleration."},{"text":"area","correct":false,"explanation":"The area under a velocity-time graph gives displacement. Acceleration is found from the graph’s gradient, which is change in velocity divided by time."},{"text":"intercept","correct":false,"explanation":"The intercept gives the velocity when the graph crosses an axis, depending on the axes used. Acceleration is represented by the gradient."},{"text":"curvature","correct":false,"explanation":"Curvature describes how the gradient changes, rather than directly giving acceleration at a point. The gradient of a velocity-time graph represents acceleration."}]},{"id":"902974e9-2169-48e9-97dc-5f1f47f05e70","cardType":"cloze","points":null,"front":"The area under a velocity-time graph represents [ ].","back":"The area under a velocity-time graph represents displacement.","mcqOptions":[{"text":"displacement","correct":true,"explanation":"The area under a velocity-time graph combines velocity with time. Since velocity multiplied by time gives displacement, the area represents displacement."},{"text":"acceleration","correct":false,"explanation":"Acceleration is represented by the gradient of a velocity-time graph, not its area. The area under the graph represents displacement."},{"text":"force","correct":false,"explanation":"Force depends on mass and acceleration and is not obtained directly from the area under a velocity-time graph. That area represents displacement."},{"text":"momentum","correct":false,"explanation":"Momentum is mass multiplied by velocity, so it cannot be found from the graph area alone. The area under a velocity-time graph represents displacement."}]},{"id":"b51d97b2-f41f-4858-ac42-b587ae47d0ff","cardType":"cloze","points":null,"front":"For an irregular velocity-time graph, the area of one square is [ ].","back":"For an irregular velocity-time graph, the area of one square is time per square $\\times$ velocity per square.","mcqOptions":[{"text":"time per square $\\times$ velocity per square","correct":true,"explanation":"Each square’s area is its horizontal width multiplied by its vertical height. On a velocity-time graph, these are time per square and velocity per square."},{"text":"time per square $\\div$ velocity per square","correct":false,"explanation":"Dividing the scale values does not calculate the geometric area of a square. Area requires width multiplied by height, so the correct expression uses multiplication."},{"text":"velocity per square $\\div$ time per square","correct":false,"explanation":"This division gives a rate related to acceleration, not the area of one graph square. The square’s area is time per square multiplied by velocity per square."},{"text":"time per square $+$ velocity per square","correct":false,"explanation":"Adding the horizontal and vertical scales does not give an area and also combines incompatible quantities. The area is found by multiplying the two scales."}]},{"id":"21b0bb02-a1ac-463b-b91a-f29e8a5aeab7","cardType":"cloze","points":null,"front":"As a falling object speeds up, [ ] increases while its weight stays constant.","back":"As a falling object speeds up, drag increases while its weight stays constant.","mcqOptions":[{"text":"drag","correct":true,"explanation":"As the object falls faster, the fluid exerts more resistance, so drag increases. The object’s weight remains constant because its mass and the gravitational field are unchanged."},{"text":"lift","correct":false,"explanation":"Lift is not the main force opposing an ordinary falling object. The force that increases with falling speed is drag, while weight remains constant."},{"text":"weight","correct":false,"explanation":"Weight does not increase simply because the object speeds up. Its weight remains constant, while drag increases as the falling speed rises."},{"text":"mass","correct":false,"explanation":"Mass is the amount of matter in the object and does not increase as it falls faster. Drag increases with speed, while the mass and weight remain constant."}]},{"id":"def8d5c2-24ce-47e2-a661-b0a4181e4502","cardType":"cloze","points":null,"front":"At terminal velocity, the velocity-time graph becomes horizontal because its [ ].","back":"At terminal velocity, the velocity-time graph becomes horizontal because its gradient is zero.","mcqOptions":[{"text":"gradient is zero","correct":true,"explanation":"A horizontal velocity-time graph has no change in velocity as time passes. Its gradient is therefore zero, indicating zero acceleration at terminal velocity."},{"text":"gradient is maximum","correct":false,"explanation":"A maximum gradient would indicate a large acceleration, meaning velocity is changing rapidly. At terminal velocity the graph is horizontal, so its gradient is zero."},{"text":"area is zero","correct":false,"explanation":"The area under the graph represents displacement and is not zero while the object is moving. The graph is horizontal because its gradient is zero."},{"text":"velocity is zero","correct":false,"explanation":"At terminal velocity the object continues moving at a constant, non-zero velocity. The graph becomes horizontal because its gradient is zero, not because velocity is zero."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"5.6.1g Acceleration","title":"5.6.1g Acceleration Flashcards","topbar":{"backHref":"/gcse/physics/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"9d0e977b-1f27-5e22-8f12-3277aba32e1a","walled":"$L38"}],"$L39","$L3a"]