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This is the definition of acceleration, so a steeper gradient represents a greater acceleration."},{"text":"Velocity","correct":false,"explanation":"Velocity is the quantity plotted on the vertical axis, not the graph’s gradient. The gradient measures change in velocity per unit time, which is acceleration."},{"text":"Displacement","correct":false,"explanation":"Displacement is obtained from the signed area under a velocity-time graph. Its gradient instead gives the rate of change of velocity, which is acceleration."},{"text":"Jerk","correct":false,"explanation":"Jerk is the rate of change of acceleration and would be the gradient of an acceleration-time graph. A velocity-time graph has acceleration as its gradient."}]},{"id":"b1ce8963-8f09-4a02-bad5-a9032e3dcdb8","cardType":"basic","points":1,"front":"What does the area under a velocity-time graph represent?","back":"Displacement","mcqOptions":[{"text":"Displacement","correct":true,"explanation":"Multiplying velocity by a small time interval gives a displacement. Summing these contributions gives the signed area under the graph, so the total area represents displacement."},{"text":"Distance","correct":false,"explanation":"This confuses velocity with speed. Negative areas reduce the net displacement, whereas distance requires the total area under a speed-time graph, with all contributions counted positively."},{"text":"Acceleration","correct":false,"explanation":"Acceleration is found from the gradient of a velocity-time graph, not its area. The signed area sums velocity multiplied by time and therefore gives displacement."},{"text":"Velocity","correct":false,"explanation":"Velocity is read directly from the vertical axis of the graph. The area combines velocity with elapsed time, producing displacement rather than another velocity."}]},{"id":"ed95f0d6-4f66-42a5-a330-62a29da2038b","cardType":"cloze","points":null,"front":"The SUVAT equations can only be used when the object's [ ] is constant and it is moving in a straight line.","back":"The SUVAT equations can only be used when the object's acceleration is constant and it is moving in a straight line.","mcqOptions":[{"text":"acceleration","correct":true,"explanation":"The SUVAT equations are derived for constant acceleration. If acceleration varies during the motion, a single value of $a$ cannot describe the whole time interval."},{"text":"velocity","correct":false,"explanation":"Constant velocity is only the special case where acceleration is zero. SUVAT can also describe changing velocity, provided the acceleration is constant and the motion is straight-line."},{"text":"displacement","correct":false,"explanation":"Displacement normally changes as an object moves, so it does not need to remain constant. The required constant quantity is acceleration, alongside motion in a straight line."},{"text":"speed","correct":false,"explanation":"Speed need not be constant during uniformly accelerated motion. SUVAT describes how velocity changes when acceleration is constant and the object moves along a straight line."}]},{"id":"fea91ef4-8851-423c-8a0e-1e9068db1955","cardType":"cloze","points":null,"front":"The SUVAT equations can only be used when the object's acceleration is [ ] and it is moving in a straight line.","back":"The SUVAT equations can only be used when the object's acceleration is constant and it is moving in a straight line.","mcqOptions":[{"text":"constant","correct":true,"explanation":"SUVAT assumes that acceleration has one unchanged value throughout the chosen time interval. This allows velocity and displacement to be related using the standard kinematic equations."},{"text":"increasing","correct":false,"explanation":"Increasing acceleration means acceleration changes with time, violating the main SUVAT assumption. The acceleration must remain constant throughout the time interval being considered."},{"text":"decreasing","correct":false,"explanation":"Decreasing acceleration is variable acceleration, so the standard SUVAT equations do not apply across the whole interval. Their derivation requires acceleration to be constant."},{"text":"zero","correct":false,"explanation":"Zero acceleration is one possible constant acceleration, but SUVAT is not restricted to that case. It applies to any constant value of acceleration, including non-zero values."}]},{"id":"0b95e5a7-85be-4a40-a6f6-ec0bc4898ac7","cardType":"cloze","points":null,"front":"The SUVAT equations can only be used when the object's acceleration is constant and it is moving in a [ ].","back":"The SUVAT equations can only be used when the object's acceleration is constant and it is moving in a straight line.","mcqOptions":[{"text":"straight line","correct":true,"explanation":"The scalar SUVAT equations assume one-dimensional motion, so velocity and acceleration can be assigned consistent positive and negative directions along a single straight line."},{"text":"circular path","correct":false,"explanation":"In circular motion the direction of velocity continuously changes, and the acceleration vector is generally not constant. Standard scalar SUVAT requires motion along a straight line."},{"text":"curved path","correct":false,"explanation":"A curved path involves changes in the direction of motion, so a single scalar direction is insufficient. Standard SUVAT is applied to motion along a straight line."},{"text":"changing direction","correct":false,"explanation":"Changing direction prevents the motion from being represented by one consistent scalar axis throughout. The stated SUVAT condition is motion in a straight line with constant acceleration."}]},{"id":"53979579-30fa-46a4-ab92-d8893370ced1","cardType":"basic","points":1,"front":"Identify the SUVAT equation used when displacement is not given and is not required.","back":"$$v = u + at$","mcqOptions":[{"text":"$$v = u + at$","correct":true,"explanation":"This equation relates initial velocity, final velocity, acceleration and time. It contains no displacement term, so it is selected when $s$ is neither known nor required."},{"text":"$$s = ut + \\frac{1}{2}at^2$","correct":false,"explanation":"This equation omits final velocity, not displacement, and explicitly contains $s$. When displacement is not involved, the appropriate equation is $v = u + at$."},{"text":"$$v^2 = u^2 + 2as$","correct":false,"explanation":"This is the time-independent SUVAT equation, but it includes displacement $s$. The equation that omits displacement is $v = u + at$."},{"text":"$$s = \\frac{u + v}{2}t$","correct":false,"explanation":"This average-velocity equation includes displacement $s$ and omits acceleration. When displacement is not given or required, use $v = u + at$ instead."}]},{"id":"fbb19bd1-18e2-462d-ad43-8b0a824426c0","cardType":"basic","points":1,"front":"Identify the SUVAT equation used when final velocity is not given and is not required.","back":"$$s = ut + \\frac{1}{2}at^2$","mcqOptions":[{"text":"$$s = ut + \\frac{1}{2}at^2$","correct":true,"explanation":"This equation relates displacement, initial velocity, acceleration and time without using final velocity. It is therefore appropriate when $v$ is neither known nor required."},{"text":"$$v = u + at$","correct":false,"explanation":"This equation contains final velocity $v$ and omits displacement. When final velocity is not involved, use $s = ut + \\frac{1}{2}at^2$."},{"text":"$$s = \\frac{u + v}{2}t$","correct":false,"explanation":"This equation calculates displacement using average velocity, but it requires final velocity $v$. The required equation is $s = ut + \\frac{1}{2}at^2$."},{"text":"$$s = vt - \\frac{1}{2}at^2$","correct":false,"explanation":"This alternative displacement equation uses final velocity $v$ and omits initial velocity. Without final velocity, use $s = ut + \\frac{1}{2}at^2$."}]},{"id":"87b5b960-45f4-4d3d-96bd-222417dd5f9e","cardType":"basic","points":1,"front":"Identify the SUVAT equation used when time is not given and is not required.","back":"$$v^2 = u^2 + 2as$","mcqOptions":[{"text":"$$v^2 = u^2 + 2as$","correct":true,"explanation":"This equation relates initial velocity, final velocity, acceleration and displacement without containing time. It is used when $t$ is neither known nor required."},{"text":"$$v = u + at$","correct":false,"explanation":"This equation directly includes time $t$ and omits displacement. When time is not involved, use the time-independent equation $v^2 = u^2 + 2as$."},{"text":"$$s = ut + \\frac{1}{2}at^2$","correct":false,"explanation":"This displacement equation contains time in two terms and omits final velocity. The SUVAT equation that eliminates time is $v^2 = u^2 + 2as$."},{"text":"$$s = \\frac{u + v}{2}t$","correct":false,"explanation":"This equation multiplies average velocity by time, so it cannot be used without $t$. The time-independent SUVAT relationship is $v^2 = u^2 + 2as$."}]},{"id":"1aed5546-53c7-499a-be40-357846ccac90","cardType":"basic","points":1,"front":"Identify the SUVAT equation used when acceleration is not given and is not required.","back":"$$s = \\frac{1}{2}(u + v)t$","mcqOptions":[{"text":"$$s = \\frac{1}{2}(u + v)t$","correct":true,"explanation":"This equation finds displacement from initial velocity, final velocity and time. Acceleration is absent, so it is the appropriate SUVAT equation when acceleration is neither known nor required."},{"text":"$$v = u + at$","correct":false,"explanation":"This equation explicitly contains acceleration and is used to relate the change in velocity to time. The equation that excludes acceleration is $s = \\frac{1}{2}(u + v)t$."},{"text":"$$s = ut + \\frac{1}{2}at^2$","correct":false,"explanation":"This equation requires acceleration and does not use final velocity. When acceleration is not given or required, use average velocity to obtain $s = \\frac{1}{2}(u + v)t$."},{"text":"$$v^2 = u^2 + 2as$","correct":false,"explanation":"This equation eliminates time rather than acceleration, so it cannot be used without knowing $a$. The acceleration-free relationship is $s = \\frac{1}{2}(u + v)t$."}]},{"id":"14ddd180-b0bc-44ab-a5b6-7cb1038b64f5","cardType":"cloze","points":null,"front":"In the absence of air resistance, a feather and a steel ball fall with [ ].","back":"In the absence of air resistance, a feather and a steel ball fall with the same acceleration.","mcqOptions":[{"text":"the same acceleration","correct":true,"explanation":"Gravitational acceleration is independent of an object’s mass when air resistance is absent. Both objects therefore accelerate towards Earth at approximately $g$, despite having very different masses."},{"text":"different accelerations","correct":false,"explanation":"Greater mass does not produce greater gravitational acceleration because gravitational force and inertia both increase with mass. Without air resistance, the correct completion is “the same acceleration”."},{"text":"accelerations proportional to their masses","correct":false,"explanation":"Although the steel ball experiences a larger gravitational force, it also has proportionally greater inertia. These effects cancel, so the correct completion is “the same acceleration”."},{"text":"the same terminal velocity","correct":false,"explanation":"Terminal velocity results from air resistance balancing weight, so it cannot occur when air resistance is absent. The feather and ball instead fall with the same acceleration."}]},{"id":"bcc60f7b-5ac8-4807-8c1c-70e66bc98383","cardType":"basic","points":1,"front":"If upward is chosen as the positive direction, what is the value of acceleration for an object in free fall near Earth?","back":"$$-9.81\\text{ m s}^{-2}$","mcqOptions":[{"text":"$$-9.81\\text{ m s}^{-2}$","correct":true,"explanation":"Gravity acts downwards, opposite to the chosen positive upward direction. Its acceleration must therefore have a negative sign, giving approximately $-9.81\\text{ m s}^{-2}$ near Earth."},{"text":"$$+9.81\\text{ m s}^{-2}$","correct":false,"explanation":"This sign would apply if downward had been chosen as positive. With upward positive, gravitational acceleration points in the negative direction and is $-9.81\\text{ m s}^{-2}$."},{"text":"$$-4.91\\text{ m s}^{-2}$","correct":false,"explanation":"This incorrectly halves gravitational acceleration, often by confusing $g$ with the coefficient $\\frac{1}{2}g$ in a displacement equation. The acceleration itself is $-9.81\\text{ m s}^{-2}$."},{"text":"$$0\\text{ m s}^{-2}$","correct":false,"explanation":"An object in free fall still accelerates due to gravity, even if its velocity is momentarily zero at the top of its path. The acceleration is $-9.81\\text{ m s}^{-2}$."}]},{"id":"6166a715-81f4-4954-8eb5-0ff8264a6c1c","cardType":"basic","points":2,"front":"How is speed calculated using a light gate and interrupt card?","back":"$$v = \\frac{L}{\\Delta t}$ where $L$ is the length of the card and $\\Delta t$ is the time the beam is blocked.","mcqOptions":[{"text":"$$v = \\frac{L}{\\Delta t}$ where $L$ is the length of the card and $\\Delta t$ is the time the beam is blocked.","correct":true,"explanation":"While the card blocks the beam, it travels a distance equal to its length. Dividing this distance by the blocking time gives the card’s average speed through the gate."},{"text":"$$v = \\frac{\\Delta t}{L}$ where $L$ is the length of the card and $\\Delta t$ is the time the beam is blocked.","correct":false,"explanation":"This reverses the speed formula and produces units of time per distance rather than distance per time. The correct calculation is $v = \\frac{L}{\\Delta t}$."},{"text":"$$v = \\frac{2L}{\\Delta t}$ where $L$ is the length of the card and $\\Delta t$ is the time the beam is blocked.","correct":false,"explanation":"This incorrectly doubles the distance travelled while the beam is blocked. The relevant distance is one card length, so the correct relationship is $v = \\frac{L}{\\Delta t}$."},{"text":"$$v = \\frac{L}{\\Delta t}$ where $L$ is the distance to the light gate and $\\Delta t$ is the time the beam is blocked.","correct":false,"explanation":"The blocking time corresponds to the interrupt card passing through the beam, not the journey to the gate. Therefore $L$ must be the card’s length in $v = \\frac{L}{\\Delta t}$."}]},{"id":"62827ac6-f37f-4986-b2c7-eef131938ceb","cardType":"cloze","points":null,"front":"In a ticker timer experiment, an increasing spacing between dots indicates the object is [ ] (speeding up).","back":"In a ticker timer experiment, an increasing spacing between dots indicates the object is accelerating (speeding up).","mcqOptions":[{"text":"accelerating","correct":true,"explanation":"A ticker timer makes dots at equal time intervals. Increasing distances between successive dots therefore show that more distance is covered per interval, meaning the object is accelerating."},{"text":"decelerating","correct":false,"explanation":"Deceleration would normally produce decreasing spacing because the object covers less distance during each equal time interval. Increasing spacing shows that the object is accelerating."},{"text":"moving at constant speed","correct":false,"explanation":"Constant speed would produce equal spacing between successive dots because equal distances would be travelled in equal times. Increasing spacing instead shows that the object is accelerating."},{"text":"stationary","correct":false,"explanation":"A stationary object would repeatedly mark the same position rather than produce increasingly separated dots. Increasing spacing means its speed is rising, so it is accelerating."}]},{"id":"daa6eb4f-5b0c-4d60-b68b-21c8f667e17d","cardType":"basic","points":1,"front":"Why is electronic timing (like light gates) preferred over manual stopwatches in motion experiments?","back":"It eliminates human reaction-time error.","mcqOptions":[{"text":"It eliminates human reaction-time error.","correct":true,"explanation":"A light gate starts and stops timing automatically as the beam is interrupted. This removes the variable delay caused by a person reacting when operating a manual stopwatch."},{"text":"It eliminates parallax error.","correct":false,"explanation":"Parallax arises when a scale is viewed from an angle and is not the main limitation of manual timing. Electronic timing specifically eliminates human reaction-time error."},{"text":"It eliminates calibration error.","correct":false,"explanation":"Electronic sensors can still be incorrectly calibrated, so this source of systematic error is not automatically removed. Their main advantage is eliminating human reaction-time error."},{"text":"It eliminates distance-measurement error.","correct":false,"explanation":"Lengths such as the interrupt card’s dimensions must still be measured and retain uncertainty. Electronic timing instead eliminates the human reaction-time error associated with starting and stopping a stopwatch."}]},{"id":"72637470-28a4-468b-bd33-aea8a476b5dc","cardType":"basic","points":2,"front":"When determining $g$ by dropping a ball, what should be plotted on the axes to get a gradient of $\\frac{g}{2}$?","back":"

y-axis: displacement ($s$)
x-axis: time squared ($t^2$)

","mcqOptions":[{"text":"

y-axis: displacement ($s$)
x-axis: time squared ($t^2$)

","correct":true,"explanation":"For release from rest, $s = \\frac{1}{2}gt^2$. Plotting $s$ against $t^2$ therefore gives a straight line whose gradient is the coefficient $\\frac{g}{2}$."},{"text":"

y-axis: time squared ($t^2$)
x-axis: displacement ($s$)

","correct":false,"explanation":"Swapping the axes makes the gradient $\\frac{2}{g}$ rather than $\\frac{g}{2}$. Displacement must be on the y-axis and time squared on the x-axis."},{"text":"

y-axis: displacement ($s$)
x-axis: time ($t$)

","correct":false,"explanation":"Displacement is proportional to $t^2$, not $t$, for an object released from rest under constant gravity. Plotting $s$ against $t^2$ gives the required gradient $\\frac{g}{2}$."},{"text":"

y-axis: velocity ($v$)
x-axis: time ($t$)

","correct":false,"explanation":"A velocity-time graph for free fall has gradient $g$, not $\\frac{g}{2}$. The required gradient comes from plotting displacement against time squared using $s = \\frac{1}{2}gt^2$."}]},{"id":"c2843db5-c7d3-48da-95e6-92bbec1cb5ed","cardType":"basic","points":1,"front":"What two components must be added together to calculate the total stopping distance?","back":"Thinking distance + braking distance","mcqOptions":[{"text":"Thinking distance + braking distance","correct":true,"explanation":"The vehicle travels during the driver’s reaction time and then continues travelling while the brakes slow it down. These thinking and braking distances together form the total stopping distance."},{"text":"Thinking time + braking distance","correct":false,"explanation":"A time cannot be added to a distance because the quantities have different units. Total stopping distance is the sum of thinking distance and braking distance."},{"text":"Thinking distance + braking time","correct":false,"explanation":"Braking time is a duration rather than a distance, so it cannot form part of a total distance. The required components are thinking distance and braking distance."},{"text":"Reaction time + braking time","correct":false,"explanation":"These two quantities would give a total time, not the distance travelled before stopping. Total stopping distance equals thinking distance plus braking distance."}]},{"id":"09ec11b6-6074-4ee7-a203-4fc0d2c90b85","cardType":"basic","points":2,"front":"List two factors that specifically increase thinking distance.","back":"

Tiredness
Alcohol or drugs
Distractions

","mcqOptions":[{"text":"

Tiredness
Alcohol or drugs
Distractions

","correct":true,"explanation":"Tiredness, alcohol or drugs, and distractions can delay a driver’s response, giving the vehicle more time to travel before braking begins."},{"text":"

Icy or wet roads
Worn tyres/brakes
Reduced friction

","correct":false,"explanation":"These conditions reduce tyre-road friction or braking effectiveness, so they increase braking distance rather than the distance travelled during the driver’s reaction time."},{"text":"

Steep downhill slope
Worn brake pads
Low tyre pressure

","correct":false,"explanation":"These are factors affecting how far the vehicle travels while stopping, not specifically how quickly the driver notices and responds to a hazard."},{"text":"

Heavy vehicle load
Worn tyres
Hot brakes

","correct":false,"explanation":"These mainly alter the forces and energy involved in stopping the vehicle. Thinking distance is instead increased by delayed reactions or greater speed during reaction time."}]},{"id":"fd9faf0d-c794-4c10-8898-7cec9d7ff1df","cardType":"basic","points":2,"front":"List two factors that specifically increase braking distance.","back":"

Icy or wet roads
Worn tyres/brakes
Higher initial speed

","mcqOptions":[{"text":"

Icy or wet roads
Worn tyres/brakes
Higher initial speed

","correct":true,"explanation":"Icy or wet roads, worn tyres or brakes, and higher initial speed all make the vehicle travel farther while its brakes bring it to rest."},{"text":"

Tiredness
Alcohol or drugs
Distractions

","correct":false,"explanation":"These factors delay the driver’s response and therefore increase thinking distance. They do not directly reduce braking forces or increase the distance needed to stop."},{"text":"

Poor concentration
Slow reactions
Fatigue

","correct":false,"explanation":"These are reaction-related influences, so they affect thinking distance. Braking distance depends on speed, friction, braking condition, and the vehicle’s kinetic energy."},{"text":"

Using a mobile phone
Illness
Lack of sleep

","correct":false,"explanation":"These conditions can make a driver respond later, but they do not specifically change the braking process after the brakes have been applied."}]},{"id":"3bbba071-6524-451e-b92c-e76ec09338d4","cardType":"cloze","points":null,"front":"For a fixed reaction time, thinking distance is proportional to [ ], whereas braking distance is proportional to speed squared.","back":"For a fixed reaction time, thinking distance is proportional to speed, whereas braking distance is proportional to speed squared.","mcqOptions":[{"text":"speed","correct":true,"explanation":"With reaction time fixed, distance equals speed multiplied by time, so thinking distance increases in direct proportion to speed."},{"text":"time","correct":false,"explanation":"Thinking distance is a distance, not directly proportional to time alone when reaction time is fixed. Speed is the remaining changing quantity in the relationship."},{"text":"acceleration","correct":false,"explanation":"Acceleration is not the proportionality for thinking distance. During a fixed reaction time, the distance travelled is determined by speed multiplied by that time."},{"text":"distance","correct":false,"explanation":"Distance is the quantity being calculated, not the factor that determines thinking distance in this fixed-reaction-time relationship; speed is proportional to it."}]},{"id":"9ccbfc70-89e3-452b-96cd-8f3006ae153f","cardType":"cloze","points":null,"front":"For a fixed reaction time, thinking distance is proportional to speed, whereas braking distance is proportional to [ ].","back":"For a fixed reaction time, thinking distance is proportional to speed, whereas braking distance is proportional to speed squared.","mcqOptions":[{"text":"speed squared","correct":true,"explanation":"Braking distance is proportional to the square of speed because kinetic energy increases with speed squared, requiring more distance to dissipate."},{"text":"speed","correct":false,"explanation":"Speed alone describes the proportionality of thinking distance, not braking distance. Braking distance depends on speed squared because kinetic energy has a squared-speed term."},{"text":"acceleration","correct":false,"explanation":"Braking distance is not directly proportional to acceleration. The speed-squared relationship follows from the vehicle’s kinetic energy and approximately constant braking force."},{"text":"time","correct":false,"explanation":"Time may affect motion, but the stated braking-distance relationship is specifically with speed squared, not reaction time or stopping time alone."}]},{"id":"daf397ba-be01-4352-832e-603557912bb7","cardType":"basic","points":1,"front":"What is the final velocity ($v$) of a ball at the highest point of its trajectory when thrown vertically upwards?","back":"$$0\\text{ m s}^{-1}$","mcqOptions":[{"text":"$$0\\text{ m s}^{-1}$","correct":true,"explanation":"At the highest point, the ball momentarily stops moving upward before reversing direction, so its instantaneous velocity is zero."},{"text":"$$9.8\\text{ m s}^{-1}$","correct":false,"explanation":"This confuses acceleration due to gravity with velocity. Gravity still gives an acceleration of about $9.8\\text{ m s}^{-2}$ at the highest point, but velocity is zero."},{"text":"$$-9.8\\text{ m s}^{-1}$","correct":false,"explanation":"The negative sign would indicate downward velocity, which develops after the highest point. Exactly at the turning point, instantaneous velocity is zero."},{"text":"$$20\\text{ m s}^{-1}$","correct":false,"explanation":"This could be an initial or earlier velocity, but the upward velocity decreases to zero at the top before the ball begins descending."}]},{"id":"643b5795-8f47-4dfd-97da-136fa2f0b530","cardType":"cloze","points":null,"front":"Free fall is defined as motion under the influence of [ ] alone, with no air resistance.","back":"Free fall is defined as motion under the influence of gravity alone, with no air resistance.","mcqOptions":[{"text":"gravity","correct":true,"explanation":"Free fall means gravity is the only force influencing the motion; air resistance and other forces are excluded from the definition."},{"text":"air resistance","correct":false,"explanation":"Air resistance is specifically excluded in the stated definition, so it cannot be the force acting alone in free fall."},{"text":"the normal reaction","correct":false,"explanation":"The normal contact force requires contact with a surface, whereas free-falling objects are acted on by gravity alone in this idealised definition."},{"text":"thrust","correct":false,"explanation":"Thrust is a propulsion force produced by an engine or similar mechanism, so motion under thrust alone is not free fall."}]},{"id":"5dfa1018-6da6-4b7d-bf5d-a3aa6306aba1","cardType":"cloze","points":null,"front":"Free fall is defined as motion under the influence of gravity alone, with no [ ].","back":"Free fall is defined as motion under the influence of gravity alone, with no air resistance.","mcqOptions":[{"text":"air resistance","correct":true,"explanation":"The ideal definition of free fall neglects air resistance, leaving gravity as the only force acting on the object."},{"text":"gravity","correct":false,"explanation":"Gravity is the force retained in the definition, not the force omitted. Free fall is motion under gravity alone with no air resistance."},{"text":"mass","correct":false,"explanation":"Mass is an intrinsic property of the object and is not a resisting force. It is not the missing condition in the definition of free fall."},{"text":"normal reaction","correct":false,"explanation":"The normal reaction acts when an object contacts a surface; it is not the force removed when defining ideal free fall."}]},{"id":"8da25027-5068-4d27-88c5-7a6e2644d1ec","cardType":"basic","points":2,"front":"What is the difference between reaction time and thinking distance?","back":"

Reaction time is a measurement of time.
Thinking distance is the displacement during that time.

","mcqOptions":[{"text":"

Reaction time is a measurement of time.
Thinking distance is the displacement during that time.

","correct":true,"explanation":"Reaction time is measured in seconds, whereas thinking distance is the displacement covered during that interval before braking starts."},{"text":"

Reaction time is a measurement of displacement.
Thinking distance is the time during that displacement.

","correct":false,"explanation":"This reverses the quantities: reaction time is temporal, while thinking distance describes displacement. The driver’s response interval is not itself a distance."},{"text":"

Reaction time is a measurement of time.
Thinking distance is a measurement of time.

","correct":false,"explanation":"Both terms have been treated as measurements of time, but thinking distance is a length or displacement travelled during the reaction time."},{"text":"

Reaction time is a measurement of distance.
Thinking distance is the displacement during that time.

","correct":false,"explanation":"Both terms have been treated as distances, but reaction time measures a duration. Only the distance travelled during that duration is a displacement."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Linear motion","title":"Linear motion Flashcards","topbar":{"backHref":"/a-level/physics/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"60d18ca6-89e6-5d19-8e12-552292379f72","walled":"$L39"}],"$L3a","$L3b"]