5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
On a winter morning the air temperature is −5 °C, and a cart rolling along a straight track has velocity component vx = −5 m/s, where the +x direction points east. Which statement is correct?
Answer and reasoning
AThe temperature is a vector; its minus sign gives it a direction, as for vx. A student who thinks any quantity that can be negative is a vector picks this. A temperature has no direction in space; its minus sign only places it below 0 °C on the scale.
BThe cart's speed is −5 m/s, since speed and velocity are the same quantity. A student who treats speed and velocity as the same quantity picks this. Speed is a scalar, the magnitude of the velocity, and is never negative: the cart's speed is 5 m/s.
CThe cart must be west of the origin, because its velocity component is negative. A student who reads the sign of a velocity component as the side of the origin picks this. A negative vx means the cart is moving west; it can be east or west of the origin.
DThe temperature is a scalar; its minus sign shows only a value below 0 °C.Correct Temperature has no direction in space, so it is a scalar. Its minus sign places the value below the zero of the Celsius scale; it does not point anywhere. The cart's −5 m/s, by contrast, is a velocity component whose sign does show a direction: west.
Working Temperature has no direction in space: scalar; −5 °C is a value below the zero of the scale. vx = −5 m/s: velocity component, sign gives direction (west); speed = |vx| = 5 m/s; the sign of vx says nothing about the cart's position.
The figure shows vectors A⃗ and B⃗ on the left grid and four arrows, P, Q, R and S, on the right grid; both grids have the same scale. Which arrow represents A⃗ + B⃗?
Answer and reasoning
AArrow P A student who draws A⃗ and B⃗ from one point and joins their heads picks this: arrow P (5 right, 2 down) is A⃗ − B⃗. The sum needs B⃗'s tail at A⃗'s head, which gives 3 right and 4 up.
BArrow QCorrect Placing the tail of B⃗ at the head of A⃗, the resultant runs from the tail of A⃗ to the head of B⃗: (4 − 1) squares right and (1 + 3) squares up, 3 right and 4 up. Arrow Q is that vector.
CArrow R A student who closes the head-to-tail chain by drawing back to the start picks this: arrow R (3 left, 4 down) is −(A⃗ + B⃗). The resultant points from the first tail to the last head, 3 right and 4 up.
DArrow S A student who ignores the leftward direction of B⃗'s horizontal part, adding it as if it pointed right, picks this: 4 + 1 = 5 right and 4 up is arrow S. B⃗ goes 1 square left, so the sum goes 4 − 1 = 3 squares right.
Working Reading the left grid: A⃗ is 4 squares right and 1 up; B⃗ is 1 square left and 3 up. Placing B⃗'s tail at A⃗'s head, the resultant from A⃗'s tail to B⃗'s head is (4 − 1) right and (1 + 3) up = 3 right, 4 up.
Aposition, speed, acceleration A student who treats speed as the same quantity as velocity picks this. Speed is the magnitude of velocity, a scalar with no direction.
Bdistance, velocity, acceleration A student who treats distance as the same quantity as displacement picks this. Distance is the length of the path traveled, a scalar with no direction.
Ctemperature, position, velocity A student who thinks a quantity that can be negative must be a vector picks this. Temperature can be below zero, but it has no direction in space, so it is a scalar.
Ddisplacement, velocity, accelerationCorrect Displacement, velocity and acceleration each have a magnitude and a direction, so all three are vectors.
Working Vectors: position, displacement, velocity, acceleration. Scalars: distance, speed, temperature.
The position vector of a point is r⃗ = (3.0î + 4.0ĵ) m, and r̂ is the unit vector in the direction of r⃗. Which statement about r̂ is correct?
Answer and reasoning
AIt is r⃗ itself, with the same magnitude, unit and direction as r⃗. A student who treats a vector and its unit vector as interchangeable picks this. r̂ = r⃗/|r⃗| keeps only the direction: its magnitude is 1, not 5.0 m.
BIt has magnitude 1 m, so its unit is the meter, like that of r⃗. A student who takes 'unit vector' to mean a vector one meter long picks this. Dividing r⃗ by its magnitude cancels the meter, so r̂ has magnitude 1 and no unit.
CIt has magnitude 1, has no unit, and points the same way as r⃗.Correct r̂ = r⃗/|r⃗| = (3.0î + 4.0ĵ) m/(5.0 m) = 0.60î + 0.80ĵ. Dividing by the magnitude leaves magnitude 1 and cancels the meter, and the direction is that of r⃗.
DIt points along x or y, whichever axis lies closer to r⃗'s direction. A student who thinks the only unit vectors are î, ĵ and k̂ picks this. r̂ points exactly along r⃗, at an angle between the axes here: r̂ = 0.60î + 0.80ĵ.
Working r̂ = r⃗/|r⃗| = (3.0î + 4.0ĵ) m/(5.0 m) = 0.60î + 0.80ĵ: magnitude 1, no unit, same direction as r⃗.
On a straight east–west road, an x-axis is chosen with its origin at a signpost and the +x direction pointing west. Car P, east of the signpost, moves east at 4 m/s. Car Q, also east of the signpost, moves west at 6 m/s. Which gives the cars' velocity components?
Answer and reasoning
AvPx = −4 m/s and vQx = +6 m/sCorrect With +x pointing west, eastward motion is negative and westward motion is positive: vPx = −4 m/s and vQx = +6 m/s. The cars' positions east of the signpost do not affect the signs of their velocities.
BvPx = +4 m/s and vQx = −6 m/s A student who treats east as positive out of habit picks this. The axis here points west, so eastward motion (car P) is negative and westward motion (car Q) is positive.
CvPx = +4 m/s and vQx = +6 m/s A student who gives the speeds as the velocity components picks this. A velocity component carries a sign for direction, and the cars move in opposite directions, so their components have opposite signs.
DvPx = −4 m/s and vQx = −6 m/s A student who takes the sign of a velocity component from the side of the origin the car is on picks this: both cars are east of the signpost, at negative x. Car Q moves west, the +x direction, so vQx is positive.
Working +x points west. Car P moves east: vPx = −4 m/s. Car Q moves west: vQx = +6 m/s. Positions (both at negative x) do not set the signs of the velocities.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.1.A.1 Scalar Fix
Scalar
A quantity described completely by a magnitude (a number with a unit), such as distance, speed, time or mass. A scalar can carry a sign on its scale (a temperature of −5 °C), but the sign does not indicate a direction in space.
Vector
A quantity described by both a magnitude and a direction, such as displacement or velocity. Two vectors are equal only if their magnitudes and their directions are both the same.
Magnitude of a vector, |A⃗| or A
The size of a vector, a non-negative number with the vector's unit. It carries no information about direction.
Students often think Any quantity that can have a minus sign is a vector, because the minus sign shows a direction. In fact No. A scalar can take negative values on its scale, as a temperature of −5 °C does. Its minus sign marks a value below the zero of the scale, not a direction in space; only vectors have directions.
1.1.A.2 Arrow representation of a vector Fix
Arrow representation of a vector
An arrow that points in the vector's direction and whose length is proportional to the vector's magnitude on a chosen scale. Moving an arrow without changing its length or direction does not change the vector it represents.
Head-to-tail addition
A graphical way to add vectors: draw the second arrow starting at the head of the first; the resultant runs from the tail of the first arrow to the head of the second.
Students often think The sum of two vectors is the arrow that joins their heads when the two arrows are drawn from the same point. In fact No. With the arrows tail to tail, the arrow from the head of B⃗ to the head of A⃗ is A⃗ − B⃗. To add, place the tail of B⃗ at the head of A⃗; the sum runs from the tail of A⃗ to the head of B⃗ (the diagonal of the parallelogram).
Students often think When vectors are placed head to tail, the resultant is the arrow that completes the loop, pointing from the head of the last vector back to the tail of the first. In fact No. The resultant runs from the tail of the first vector to the head of the last one. An arrow from the last head back to the first tail is the negative of the resultant: added to the chain, it would give zero.
1.1.A.3 Distance Fix
Distance
The total length of the path an object travels; a scalar, never negative. SI unit: meter (m).
Speed
A scalar: the rate at which an object covers distance. At an instant it equals the magnitude of the velocity. SI unit: m/s.
Position, r⃗ (or x in one dimension)
A vector giving an object's location relative to a chosen origin. SI unit: m.
Displacement
A vector: the change in an object's position, from its initial position to its final position. Its magnitude is not greater than the distance traveled. SI unit: m.
Velocity and acceleration as vectors
Velocity (SI unit m/s) and acceleration (SI unit m/s²) each have a magnitude and a direction; their directions need not be the same.
Students often think Speed and velocity are the same quantity under two names, so a speed can be negative and a speed has a direction, just as a velocity does. In fact No. Velocity is a vector, so a velocity component carries a sign that gives its direction. Speed is a scalar: the magnitude of the velocity, never negative. A velocity component of −5 m/s means a speed of 5 m/s.
Students often think Distance and displacement are the same quantity, so the distance traveled always equals the magnitude of the displacement (and a displacement of zero means a distance of zero). In fact No. Distance is the total length of the path; the magnitude of the displacement is the straight-line separation of the start and end points. They are equal only for motion along a straight line in one direction; otherwise the distance is greater.
1.1.A.4 Magnitude–direction form and unit vector form Fix
Magnitude–direction form and unit vector form
Two equivalent ways to state a vector: a magnitude with a direction (for example, 5.0 m at 53° above the +x-axis), or components along the axes (for example, (3.0î + 4.0ĵ) m). For a vector of magnitude A at angle θ from the +x-axis, Ax = A cos θ and Ay = A sin θ; the sine and cosine must be matched to the axis from which θ is measured.
Unit vectors î, ĵ, k̂
Vectors of magnitude 1, with no unit, pointing in the +x, +y and +z directions. In A⃗ = Ax î + Ay ĵ + Az k̂, the signed components Ax, Ay and Az carry the vector's unit.
Component of a vector
The signed number that multiplies a unit vector in unit vector notation. Its sign shows whether the vector points toward the positive or negative end of that axis. The magnitude of a vector is √(Ax² + Ay² + Az²), not the sum of its components.
Position vector r⃗ and unit vector r̂
r⃗ points from the origin to a point. r̂ = r⃗/|r⃗| has magnitude 1, has no unit, and points in the same direction as r⃗.
Resultant vector
The vector sum of two or more vectors. Its components are the sums of the corresponding components: for C⃗ = A⃗ + B⃗, Cx = Ax + Bx and Cy = Ay + By.
Vector subtraction
A⃗ − B⃗ = A⃗ + (−B⃗): reverse B⃗ and add. In components, (A⃗ − B⃗)x = Ax − Bx and (A⃗ − B⃗)y = Ay − By.
Students often think When two vectors are combined, their magnitudes are subtracted, so the magnitude of a sum (or difference) of vectors is the difference of their magnitudes. In fact Not in general. Only for vectors pointing in exactly opposite directions is the magnitude of their sum the difference of their magnitudes. In general, magnitudes must be found from components: |A⃗ − B⃗| = √((Ax − Bx)² + (Ay − By)²).
Students often think The magnitude of a sum of vectors is the sum of their magnitudes, whatever their directions. In fact Only if the vectors point in the same direction. In general |A⃗ + B⃗| is less than |A⃗| + |B⃗|: for perpendicular vectors it is √(A² + B²), and for opposite vectors it is |A − B|.
1.1.A.5 Sign convention in one dimension Fix
Sign convention in one dimension
On a line, one direction is chosen as positive; a vector component pointing the other way is negative. The sign shows direction only: −8 m/s and +8 m/s have the same magnitude.
Students often think The sign of an object's velocity component tells which side of the origin it is on (or the sign of its position tells which way it is moving). In fact No. The sign of a velocity component gives the direction of motion; the sign of the position gives the side of the origin the object is on. An object at x = +3 m can have vx = −2 m/s.
Students often think A negative vector or component is smaller than a positive one: −8 m/s is less than +5 m/s, a magnitude can be negative, and a negative component reduces the size of a vector. In fact No. The magnitude of a vector is never negative, and a minus sign on a component shows direction, not size. A component of −8 m/s has a greater magnitude than one of +5 m/s, and (−2)² adds to a sum of squares just as (+2)² does.
10 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 10
A student claims that two displacements, each of magnitude 5 m, can add to give a total displacement of magnitude zero. Which reasoning correctly supports the claim?
Answer and reasoning
AA person can walk 5 m and back again, which makes the total distance zero. A student who treats distance and displacement as the same quantity picks this. Walking 5 m and back covers a distance of 10 m; it is the displacement, not the distance, that is zero.
BVector magnitudes combine by subtraction, so 5 m − 5 m gives a total of zero. A student who thinks magnitudes are always subtracted when vectors combine picks this. Two 5 m displacements in the same direction add to 10 m; the sum is zero only because the directions are opposite.
CDisplacements are vectors, so two of 5 m in opposite directions add to zero.Correct A displacement has a direction as well as a magnitude. Two displacements of 5 m pointing in opposite directions, such as 5 m east and 5 m west, add to a zero vector, so the claim is supported.
DA backward step has a magnitude of −5 m, and 5 m + (−5 m) gives zero. A student who thinks a magnitude can be negative picks this. Magnitudes are never negative: both displacements have magnitude 5 m, and it is their opposite directions (opposite signs of their components) that make them cancel.
Working Displacements are vectors: 5 m east + 5 m west = (+5 m) + (−5 m) = 0. Each magnitude is 5 m; the cancellation comes from the opposite directions.
The figure shows three pairs of vectors, X, Y and Z, drawn to the same scale; the number beside each arrow is its magnitude in units. Which ranking of the magnitudes of the vector sums of the three pairs is correct?
Answer and reasoning
AX > Y > ZCorrect Arrows in the same direction add fully (7 units); perpendicular arrows give √(4² + 3²) = 5 units; arrows in opposite directions partly cancel (1 unit). So X > Y > Z: the magnitude of a sum depends on the directions, not only on the magnitudes.
BX = Y = Z A student who adds magnitudes whatever the directions picks this, getting 4 + 3 = 7 units for every pair. Only pair X points the same way; for Y the sum is 5 units and for Z it is 1 unit.
CZ > Y > X A student who draws each sum as the arrow joining the two heads picks this: that arrow is the difference of the vectors, 1 unit for X, 5 for Y and 7 for Z. Placed head to tail, the sums are 7, 5 and 1 units.
DX = Z > Y A student who ignores the direction of the leftward arrow in Z, treating it as pointing right, picks this: Z then gives 7 units like X. The leftward arrow must enter with its opposite sign, so Z gives 4 − 3 = 1 unit.
Working X: same direction, 4 + 3 = 7 units. Y: perpendicular, √(4² + 3²) = 5 units. Z: opposite directions, 4 − 3 = 1 unit. So X > Y > Z.
A runner starts at the easternmost point of a circular track of radius 20 m and runs counterclockwise, as seen from above, until she reaches the northernmost point of the track. What is the magnitude of her displacement?
Answer and reasoning
A31 m A student who equates displacement with distance picks this: the quarter-circle arc is (π/2)(20 m) ≈ 31 m long. The displacement is the straight line from start to end, which is shorter.
B28 mCorrect The displacement runs straight from the start point to the end point. With the center as origin, it goes from (20 m, 0) to (0, 20 m), so Δr⃗ = (−20î + 20ĵ) m and its magnitude is √(20² + 20²) m ≈ 28 m.
C20 m A student who takes the displacement to be the runner's final distance from a reference point, the center, picks this. Displacement is measured from where she started, not from the center.
D40 m A student who adds the sizes of the displacement's components picks this: 20 m west plus 20 m north gives 40 m. Perpendicular components combine as √(20² + 20²) m ≈ 28 m.
Working Take the center of the track as origin. Start (20 m, 0), end (0, 20 m). Δr⃗ = (−20 m)î + (20 m)ĵ, |Δr⃗| = √(20² + 20²) m = 20√2 m ≈ 28 m. (The distance along the arc is (π/2)(20 m) ≈ 31 m.)
The figure shows a displacement d⃗ that starts at point P. Which expression gives d⃗ in unit vector notation?
Answer and reasoning
A(−8.0î + 6.0ĵ) m A student who always uses cosine for the x-component picks this: (10 m)cos 37° = 8.0 m along x. Here the angle is measured from the y-direction, so the x-component is (10 m)sin 37° = 6.0 m.
B(−4.0î + 9.0ĵ) m A student who describes a vector by the coordinates of its head picks this: the head of d⃗ is at (2.0 − 6.0, 1.0 + 8.0) m = (−4.0, 9.0) m. The components of d⃗ are the changes from tail to head, −6.0 m and +8.0 m.
C(−6.0î + 8.0ĵ) mCorrect The 37° angle is measured from the +y direction, so the y-component is adjacent to it: (10 m)cos 37° = 8.0 m. The x-component is opposite the angle, (10 m)sin 37° = 6.0 m, and points in the −x direction. Where d⃗ starts does not change its components.
D(−0.6î + 0.8ĵ) m A student who writes the unit vector in place of the vector picks this: −0.6î + 0.8ĵ has magnitude 1 and gives only the direction. d⃗ is 10 m long, so its components are 10 times larger.
Working The angle of 37° is measured from the +y direction, so the component along y is adjacent to the angle: dy = (10 m)cos 37° = (10 m)(0.80) = +8.0 m. The x-component is opposite the angle and points in the −x direction: dx = −(10 m)sin 37° = −(10 m)(0.60) = −6.0 m. d⃗ = (−6.0î + 8.0ĵ) m. The starting point P does not affect the components.
A point has position vector r⃗ = (1.0î − 2.0ĵ + 2.0k̂) m. What is the magnitude of r⃗?
Answer and reasoning
A3.0 mCorrect The three components are perpendicular, so |r⃗| = √(1.0² + (−2.0)² + 2.0²) m = √9.0 m = 3.0 m. The minus sign on the y-component disappears when it is squared.
B5.0 m A student who adds the sizes of the components picks this: 1.0 + 2.0 + 2.0 = 5.0 m. Perpendicular components combine as the square root of the sum of their squares.
C2.2 m A student who uses the two-dimensional formula √(x² + y²) picks this: √(1.0² + 2.0²) m ≈ 2.2 m. The z-component must be included as well.
D1.0 m A student who thinks a negative component makes the vector smaller subtracts its square: √(1.0 − 4.0 + 4.0) m = 1.0 m. The square of −2.0 is +4.0, so every component adds to the sum of squares.
Working |r⃗| = √((1.0)² + (−2.0)² + (2.0)²) m = √(1.0 + 4.0 + 4.0) m = √9.0 m = 3.0 m.
The figure shows two displacements, A⃗ and B⃗, drawn from the same point. What is the magnitude of A⃗ + B⃗?
Answer and reasoning
A3.0 mCorrect Resolve B⃗: Bx = (5.0 m)cos 53° = +3.0 m and By = −(5.0 m)sin 53° = −4.0 m. Adding A⃗ = +4.0 m along y gives a y-component of zero, so A⃗ + B⃗ = (3.0 m)î, of magnitude 3.0 m, less than either vector.
B8.5 m A student who drops the minus sign of B⃗'s downward component picks this: (3.0 m, 4.0 + 4.0 m) gives √(3.0² + 8.0²) m ≈ 8.5 m. B⃗ points below the x-axis, so its y-component is −4.0 m and cancels A⃗.
C6.4 m A student who uses √(A² + B²) for any two vectors picks this: √(4.0² + 5.0²) m ≈ 6.4 m. That rule holds only for perpendicular vectors; here the angle between A⃗ and B⃗ is 143°.
D9.0 m A student who adds the magnitudes whatever the directions picks this: 4.0 m + 5.0 m = 9.0 m. That happens only for vectors in the same direction; these largely oppose each other.
Working A⃗ = (0, +4.0 m). B⃗: Bx = (5.0 m)cos 53° = (5.0 m)(0.60) = +3.0 m; By = −(5.0 m)sin 53° = −(5.0 m)(0.80) = −4.0 m. Sum: (3.0 m, 0), magnitude 3.0 m.
Vectors A⃗ = aî + 2aĵ and B⃗ = −3aî + aĵ, where a is a positive constant. What is the magnitude of A⃗ − B⃗?
Answer and reasoning
A2.2a A student who drops the minus sign of B⃗'s x-component, taking B⃗ = 3aî + aĵ, picks this: A⃗ − B⃗ = −2aî + aĵ, of magnitude √5 a ≈ 2.2a. The −3a must keep its sign, so subtracting it adds 3a.
B4.1aCorrect Subtract component by component: (a − (−3a))î + (2a − a)ĵ = 4aî + aĵ. Its magnitude is √(16a² + a²) = √17 a ≈ 4.1a.
C3.6a A student who adds B⃗ instead of subtracting it picks this: A⃗ + B⃗ = −2aî + 3aĵ, of magnitude √13 a ≈ 3.6a. Subtraction reverses B⃗ before adding.
D5.0a A student who finds A⃗ − B⃗ = 4aî + aĵ correctly but then adds its components picks this: 4a + a = 5a. Perpendicular components combine as √(16a² + a²) = √17 a.
Working A⃗ − B⃗ = (a − (−3a))î + (2a − a)ĵ = 4aî + aĵ. |A⃗ − B⃗| = √(16a² + a²) = √17 a ≈ 4.1a.
The figure shows the three straight legs of a walk from Start to End; each leg's length is given in terms of a distance d. What is the magnitude of the walker's displacement from Start to End?
Answer and reasoning
A8.0d A student who treats the displacement as the distance walked picks this: d + 3d + 4d = 8d. The displacement depends only on where the walk starts and ends.
B5.8d A student who ignores the direction of the westward leg, adding it as if it went east, picks this: x = d + 4d = 5d, y = 3d, magnitude √34 d ≈ 5.8d. The westward leg's x-component is −4d.
C6.0d A student who finds the components −3d and 3d correctly but adds their sizes picks this: 3d + 3d = 6d. Perpendicular components combine as √(9d² + 9d²) ≈ 4.2d.
D4.2dCorrect Add components (x east, y north): x = d − 4d = −3d and y = 3d. The displacement is −3dî + 3dĵ, of magnitude √(9d² + 9d²) = 3√2 d ≈ 4.2d.
Working With x east and y north: legs dî, 3dĵ and −4dî. Sum: (d − 4d)î + 3dĵ = −3dî + 3dĵ. Magnitude √(9d² + 9d²) = 3√2 d ≈ 4.2d.
Two carts move along a straight track that lies along an x-axis. Cart 1 has velocity component vx = −8 m/s and cart 2 has vx = +5 m/s. A student claims that cart 1 is moving more slowly than cart 2, because −8 is less than +5. Which statement correctly evaluates the claim?
Answer and reasoning
AIt is right: −8 m/s is less than +5 m/s, so cart 1 is the slower of the two. A student who ranks velocity components as numbers on a number line picks this. Speeds are compared by magnitude: 8 m/s is greater than 5 m/s.
BIt is wrong: the minus sign gives cart 1's direction; its 8 m/s speed is greater.Correct The sign of a velocity component shows direction only. Cart 1's speed is |−8 m/s| = 8 m/s, greater than cart 2's 5 m/s, so cart 1 is the faster cart; it moves in the −x direction.
CIt is wrong: the minus sign puts cart 1 left of the origin, and its speed is 8 m/s. A student who reads the sign of a velocity component as the side of the origin picks this. The minus sign gives the direction of motion; the stem says nothing about where cart 1 is.
DIt is right: the minus sign shows that cart 1 is slowing, so it is the slower one. A student who reads a negative velocity as slowing down picks this. The minus sign gives only the direction of motion; and even a slowing cart is, at this instant, moving at 8 m/s.
Working Speed = |vx|: cart 1, 8 m/s; cart 2, 5 m/s. Cart 1 is faster and moves in the −x direction.
Vectors A⃗ and B⃗ are perpendicular and have equal magnitudes. The magnitude of B⃗ is then tripled, while its direction and A⃗ stay the same. By what factor does the magnitude of A⃗ + B⃗ increase?
Answer and reasoning
A2.0 A student who adds the magnitudes picks this: A + A = 2A before and A + 3A = 4A after, a factor of 2. Perpendicular vectors add as √(A² + B²).
B3.0 A student who scales the sum by the same factor as B⃗ picks this. Only one of the two vectors changes, so the sum grows by less than 3.
C2.2Correct For perpendicular vectors |A⃗ + B⃗| = √(A² + B²). With equal magnitudes A it is √2 A; with B = 3A it is √(A² + 9A²) = √10 A. The factor is √10/√2 = √5 ≈ 2.2.
D1.7 A student who sees the square root and multiplies the result by √3 picks this. B appears squared, so tripling B makes B² nine times larger, and A² does not change: the factor is √(10/2) ≈ 2.2.
Working Let each magnitude be A. Before: |A⃗ + B⃗| = √(A² + A²) = √2 A. After: √(A² + 9A²) = √10 A. Factor √10/√2 = √5 ≈ 2.2.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
