SCI 101: Assignment 1, solved step by step

55 unique questions collected from your four files (repeated questions appear only once), grouped by topic. Each one has the correct answer, a worked solution and a note on the usual trap.

Formula sheet
Magnitude (2D / 3D)
|v| = √(x² + y²)  or  √(x² + y² + z²)
Unit vector
magnitude = 1
Dot product
A·B = AxBx + AyBy + AzBz = AB cos θ   (a number)
Cross product magnitude
|A×B| = AB sin θ   (a vector)
Perpendicular / parallel
dot product = 0 → perpendicular; cross product = 0 → parallel
Average velocity
vavg = Δx / Δt = (x2 − x1)/(t2 − t1)
Average acceleration
a = (v − v0) / t
Constant acceleration
v = v0 + at  |  x = v0t + ½at²  |  v² = v0² + 2as
Free fall
same equations with a = g = 9.8 m/s² downward

Use g = 9.8 m/s² and ignore air resistance unless a question says otherwise.

Vector basics

Magnitude, unit vectors, components, negatives and adding vectors.

Q1True / False

The magnitude of the vector v = 3i + 4j is 5.

  • True
  • False

Answer True

  1. Magnitude = square root of the sum of squared components: |v| = √(x² + y²).
  2. |v| = √(3² + 4²) = √(9 + 16) = √25 = 5.

Watch out: 3-4-5 is a famous right triangle. Spot it and you skip the calculator.

Q2True / False

The magnitude of the vector v = 2i + 5j is 5.

  • True
  • False

Answer False

  1. |v| = √(2² + 5²) = √(4 + 25) = √29 ≈ 5.39.
  2. 5.39 is not 5, so the statement is False.

Watch out: Don't just read off the biggest component. Always square and add.

Q3Multiple choice

The magnitude of the vector v = 2i + j − 9k is:

  • A9.3
  • B36
  • C6
  • D−6

Answer A. 9.3

  1. In 3D: |v| = √(x² + y² + z²).
  2. Components: 2, 1, −9. Squares: 4, 1, 81.
  3. |v| = √(4 + 1 + 81) = √86 ≈ 9.3.

Watch out: A magnitude is never negative (so D is out), and the square root of a sum of squares can't be a plain sum like 36.

Q4Multiple choice

If the velocity of a particle is given by the vector (2i + 3j − 5k) m/s, then its speed is:

  • A√38 m/s
  • B√10 m/s
  • C4 m/s
  • D10 m/s

Answer A. √38 m/s

  1. Speed is the magnitude of the velocity vector.
  2. Speed = √(2² + 3² + (−5)²) = √(4 + 9 + 25) = √38 m/s (≈ 6.16 m/s).

Watch out: Velocity is a vector; speed is its length. The minus sign disappears once squared.

Q5True / False

The vector (5, 2, 3) can be written in terms of unit vectors as 5i + 2j + 3k.

  • True
  • False

Answer True

  1. The three numbers are the x, y, z components.
  2. Each component multiplies its unit vector: x goes with i, y with j, z with k.
  3. So (5, 2, 3) = 5i + 2j + 3k. True.
Q6True / False

Any vector can be expressed as the sum of its components.

  • True
  • False

Answer True

  1. A vector A can be split along the axes: A = Axi + Ayj + Azk.
  2. This is exactly how we resolve a vector into components, so the statement is True.
Q7Multiple choice

If u = (4, 5), then 6u = ___

  • A(24, 30)
  • B(10, 11)
  • C(4, 5)
  • D(24, 6)

Answer A. (24, 30)

  1. Multiplying a vector by a number multiplies every component by that number.
  2. 6u = (6×4, 6×5) = (24, 30).

Watch out: Scalar multiplication never adds. Options B and D come from adding or only scaling one component.

Q8Multiple choice

What is the magnitude of x that makes the vector (0, 3/5, x) a unit vector? (Hint: a unit vector has magnitude 1.)

  • A4/5
  • B2/5
  • C−2/5
  • D1

Answer A. 4/5

  1. Set the magnitude equal to 1: √(0² + (3/5)² + x²) = 1.
  2. Square both sides: 9/25 + x² = 1.
  3. x² = 1 − 9/25 = 16/25, so |x| = 4/5.

Watch out: 3-4-5 again: 3/5 and 4/5 are the sides of a unit-length hypotenuse.

Q9Multiple choice

If u is a unit vector, which of the following statements is correct?

  • A−u is a unit vector with opposite direction to u
  • B−u is a unit vector with the same direction as u
  • C−u is not a unit vector
  • D−u has a zero magnitude

Answer A. −u is a unit vector with opposite direction to u

  1. The minus sign flips the direction but does not change the length.
  2. |−u| = |u| = 1, so −u is still a unit vector, pointing the opposite way.
Q10Multiple choice

What is the negative of a vector?

  • AA vector with the same magnitude but opposite direction
  • BA vector with zero magnitude
  • CA vector with the opposite direction but zero magnitude
  • DNone of the given answers

Answer A. A vector with the same magnitude but opposite direction

  1. The negative of A is −A: same length, reversed direction.
  2. Check: A + (−A) = 0, the zero vector.
Q11Multiple choice

Two vectors have the same magnitude but opposite direction. These vectors are called:

  • Aopposite
  • Bparallel
  • Cequal
  • Dperpendicular

Answer A. opposite

  1. Same magnitude and same direction = equal vectors.
  2. Same magnitude and reversed direction = opposite vectors.
  3. Perpendicular means 90° apart; parallel means same or opposite line without needing equal length.

Watch out: The question says “opposite direction”, so the name is simply “opposite”.

Q12True / False

Two vectors having the same magnitude are always equal.

  • True
  • False

Answer False

  1. Equal vectors need both the same magnitude and the same direction.
  2. Example: (3, 0) and (0, 3) both have magnitude 3, but they point in different directions. False.
Q13True / False

The sum of two unit vectors is also a unit vector.

  • True
  • False

Answer False

  1. Take i and j, both unit vectors. Their sum is (1, 1) with magnitude √2 ≈ 1.41, not 1.
  2. One counter-example is enough to make the statement False. (It would only work for the special angle of 120° between them.)
Q14Multiple choice

The resultant of two vectors depends on:

  • ABoth their magnitudes and the angle between them
  • BOnly the angle between them
  • CNeither magnitude nor direction
  • DOnly their magnitudes

Answer A. Both their magnitudes and the angle between them

  1. Resultant magnitude: R = √(A² + B² + 2AB cos θ).
  2. The formula uses both lengths (A, B) and the angle θ, so all three matter.
Q15True / False

If two vectors acting on a body keep it in equilibrium, their vector sum is zero.

  • True
  • False

Answer True

  1. Equilibrium means no net force: ΣF = 0.
  2. So the vector sum of the forces must be zero. True.

Dot product, cross product and perpendicular vectors

Computing both products, and what they tell you about the angle between vectors.

Q16Multiple choice

The dot product between u = (2, −3, 5) and v = (1, 6, 4) is equal to:

  • A4
  • B40
  • C−4
  • D−40

Answer A. 4

  1. Multiply matching components, then add: u·v = uxvx + uyvy + uzvz.
  2. = (2)(1) + (−3)(6) + (5)(4) = 2 − 18 + 20 = 4.

Watch out: The dot product gives a single number (a scalar), not a vector.

Q17True / False

The dot product of two vectors is always a vector.

  • True
  • False

Answer False

  1. Dot product = a number: A·B = AB cos θ. It is a scalar.
  2. The cross product is the one that gives a vector. So the statement is False.
Q18Multiple choice

Two vectors are perpendicular if:

  • Atheir dot product is zero
  • Btheir dot product is non-zero
  • Ctheir cross product is the zero vector
  • Dtheir cross product is non-zero

Answer A. their dot product is zero

  1. A·B = AB cos θ. At θ = 90°, cos 90° = 0, so the dot product is 0.
  2. A zero cross product means the vectors are parallel (sin θ = 0), the opposite situation.

Watch out: Dot = 0 → perpendicular. Cross = 0 → parallel.

Q19True / False

If two vectors a = (12, x) and b = (3, 6) are perpendicular, then the value of x = −6.

  • True
  • False

Answer True

  1. Perpendicular → dot product = 0.
  2. (12)(3) + (x)(6) = 0 → 36 + 6x = 0.
  3. 6x = −36 → x = −6. True.
Q20Multiple choice

Which of the following vectors is perpendicular to the vector v = (2, −4)?

  • A(4, 2)
  • B(−2, 4)
  • C(−2, −4)
  • D(−4, 2)

Answer A. (4, 2)

  1. Test each option: the perpendicular one has dot product 0 with (2, −4).
  2. (4, 2): 2(4) + (−4)(2) = 8 − 8 = 0 ✓
  3. (−2, 4): −4 − 16 = −20  |  (−2, −4): −4 + 16 = 12  |  (−4, 2): −8 − 8 = −16.

Watch out: Shortcut in 2D: swap the components and flip one sign. (2, −4) → (4, 2).

Q21Multiple choice

Find the cross product A × B where A = (3, 0, 2) and B = (−2, 5, 0).

  • A(−10, −4, 15)
  • B(10, 4, 15)
  • C(0, 3, 0)
  • D(15, 0, 10)

Answer A. (−10, −4, 15)

  1. Formula: A×B = (AyBz − AzBy,  AzBx − AxBz,  AxBy − AyBx).
  2. x: (0)(0) − (2)(5) = −10
  3. y: (2)(−2) − (3)(0) = −4
  4. z: (3)(5) − (0)(−2) = 15
  5. Result: (−10, −4, 15).

Watch out: The y-component is the one people get wrong. Notice it is Az·Bx − Ax·Bz (the order is flipped).

Q22Multiple choice

If v and u are vectors with |v| = 10, |u| = 8 and θ = 30° between them, then |v × u| =

  • A40
  • B80
  • C40√(3/2)
  • D40√3

Answer A. 40

  1. Magnitude of a cross product: |v×u| = |v||u| sin θ.
  2. = (10)(8) sin 30° = 80 × 0.5 = 40.

Watch out: Memorise sin 30° = 1/2. The dot product would use cos instead.

Q23True / False

Cross product is commutative while dot product is not commutative.

  • True
  • False

Answer False

  1. It is exactly the other way around.
  2. Dot product: A·B = B·A (commutative).
  3. Cross product: A×B = −(B×A) (anti-commutative, the order flips the direction). So the statement is False.

Concepts of motion: speed, velocity and acceleration

Conceptual questions. Focus on what is a vector (has direction) and what is not.

Q24Multiple choice

At a particular instant, the velocity of a body is called:

  • AInstantaneous Velocity
  • BInstantaneous Acceleration
  • CInstantaneous Displacement
  • DInstantaneous Speed

Answer A. Instantaneous Velocity

  1. Velocity at one specific moment is the instantaneous velocity, the limit of Δx/Δt as Δt → 0.
  2. “Instantaneous speed” is only its magnitude, but the question asks about velocity.
Q25True / False

A car travels from Dammam to Riyadh; the speedometer measures the average speed.

  • True
  • False

Answer False

  1. A speedometer shows how fast the car is going right now: the instantaneous speed.
  2. Average speed = total distance ÷ total time, which a speedometer does not show. False.
Q26True / False

Going in a straight line at the same speed is called Constant Velocity.

  • True
  • False

Answer True

  1. Velocity has magnitude and direction.
  2. Same speed (constant magnitude) + straight line (constant direction) = constant velocity. True.
Q27True / False

If the velocity of a moving particle is constant, then its average speed and instantaneous speed are the same.

  • True
  • False

Answer True

  1. Constant velocity means constant speed and no change in direction.
  2. If the speed never changes, the average speed over any interval equals the speed at every instant. True.
Q28True / False

A car drove in a straight line at a constant speed for three minutes and then made a U-turn. In this situation, the velocity remained constant.

  • True
  • False

Answer False

  1. Velocity is a vector, so a change in direction is a change in velocity, even if the speed stays the same.
  2. The U-turn reverses the direction, so velocity did not remain constant. False.
Q29True / False

If an object's speed is increasing, then the magnitude of its acceleration is not zero.

  • True
  • False

Answer True

  1. Acceleration is the rate of change of velocity: a = Δv/Δt.
  2. If speed increases, velocity changes, so a ≠ 0. True.
Q30True / False

If the speed of a particle increases in the negative direction, then its acceleration must be positive.

  • True
  • False

Answer False

  1. Speeding up means velocity and acceleration point the same way.
  2. The particle moves in the negative direction, so acceleration is negative, not positive. False.

Watch out: Speeding up: v and a same sign. Slowing down: opposite signs.

Q31True / False

An object is moving in the direction of the negative x-axis. If the object is slowing down then its acceleration must be negative.

  • True
  • False

Answer False

  1. Slowing down means acceleration points opposite to the velocity.
  2. Velocity is negative, so acceleration is positive. The statement says negative, so it is False.
Q32Multiple choice

An automobile moves on a highway at a constant speed of 80 km/h in the negative direction. Which of the following statements is true about its acceleration?

  • AThe acceleration is zero
  • BThe acceleration is positive
  • CThe acceleration is negative
  • DThe acceleration is not constant

Answer A. The acceleration is zero

  1. Constant speed in a straight line means the velocity does not change.
  2. a = Δv/Δt = 0. The direction being negative does not matter. Acceleration is zero.
Q33True / False

If an object is falling from a height of 0.5 m, then its displacement is 0.5 m downward.

  • True
  • False

Answer True

  1. Displacement is the straight-line vector from the start to the end position.
  2. It starts 0.5 m above the ground and ends at the ground: 0.5 m, pointing down. True.

Straight-line motion: calculations

Average velocity, average acceleration and the constant-acceleration equations.

Q34True / False

A runner moves along the x-axis. During a 4.0 s interval, the position changes from x1 = 30.0 m to x2 = 50.0 m. The average velocity of the runner will be 5 m/s.

  • True
  • False

Answer True

  1. Average velocity = displacement ÷ time = (x2 − x1)/Δt.
  2. = (50 − 30)/4 = 20/4 = 5 m/s. True.
Q35True / False

A runner moves along the x-axis. During a 4.0 s interval, the position changes from x1 = 20.0 m to x2 = 40.0 m. The average velocity of the runner will be 5 m/s.

  • True
  • False

Answer True

  1. Displacement = 40 − 20 = 20 m.
  2. vavg = 20/4 = 5 m/s. True.
Q36True / False

A footballer moves along the x-axis. During a 4.00 s interval, his position changes from x1 = 40.0 m to x2 = 60.0 m. His average velocity will be 4 m/s.

  • True
  • False

Answer False

  1. Displacement = 60 − 40 = 20 m.
  2. vavg = 20/4 = 5 m/s, not 4 m/s. False.
Q37Multiple choice

A particle is located at x1 = −10.0 m at t = 4 s and at x2 = 30.0 m at t = 6.00 s. Its average velocity over this interval is:

  • A20.0 m/s
  • B10.0 m/s
  • C40.0 m/s
  • D45.0 m/s

Answer A. 20.0 m/s

  1. Δx = 30 − (−10) = 40 m (subtracting a negative adds).
  2. Δt = 6 − 4 = 2 s.
  3. vavg = 40/2 = 20 m/s.

Watch out: Choice C (40) forgets to divide by the time interval.

Q38Multiple choice

A particle moves according to x(t) = 3t² − 1 (x in metres, t in seconds). Find the average velocity from t = 3 s to t = 5 s.

  • A24 m/s
  • B48 m/s
  • C12 m/s
  • D6 m/s

Answer A. 24 m/s

  1. Find the positions: x(3) = 3(9) − 1 = 26 m and x(5) = 3(25) − 1 = 74 m.
  2. Δx = 74 − 26 = 48 m, Δt = 2 s.
  3. vavg = 48/2 = 24 m/s.

Watch out: This is average velocity, so use two positions. Don't differentiate (that would give the instantaneous velocity).

Q39Multiple choice

A car travels along a straight line at a constant speed of 50.0 km/h for a distance d = 250 km. Find the time taken.

  • A5 hours
  • B4 hours
  • C6 hours
  • D10 hours

Answer A. 5 hours

  1. At constant speed: d = vt, so t = d/v.
  2. t = 250 km ÷ 50 km/h = 5 h.
Q40Multiple choice

What will be the average acceleration of a truck if it accelerates from stop to 60 m/s in 3 seconds?

  • A20 m/s²
  • B30 m/s²
  • C40 m/s²
  • D60 m/s²

Answer A. 20 m/s²

  1. From a stop means v0 = 0.
  2. a = (v − v0)/t = (60 − 0)/3 = 20 m/s².
Q41Multiple choice

An object's speed at t1 = 2 s is v(2) = 6 m/s. If its average acceleration between t1 = 2 s and t2 = 5 s is 10 m/s², what is its speed at t2 = 5 s?

  • A36 m/s
  • B18 m/s
  • C24 m/s
  • D40 m/s

Answer A. 36 m/s

  1. Rearrange a = Δv/Δt into v = v0 + aΔt.
  2. Δt = 5 − 2 = 3 s.
  3. v(5) = 6 + (10)(3) = 36 m/s.

Watch out: Don't use 5 s as the time. The acceleration acts only for the 3 s between the two instants.

Q42Multiple choice

If a particle moves from rest and accelerates at 4 m/s², how much time will it take to reach a velocity of 16 m/s?

  • A4 sec
  • B0.25 sec
  • C16 sec
  • D10 sec

Answer A. 4 sec

  1. v = v0 + at with v0 = 0.
  2. t = v/a = 16/4 = 4 s.
Q43Multiple choice

An object travels along the positive x-axis, starting from rest, at a constant acceleration of 4 m/s². How long does it take to travel 50 m?

  • A5 seconds
  • B25 seconds
  • C50 seconds
  • D10 seconds

Answer A. 5 seconds

  1. Use x = v0t + ½at² with v0 = 0.
  2. 50 = ½(4)t² = 2t² → t² = 25.
  3. t = 5 s.

Watch out: Options 25 and 50 forget to take the square root.

Q44Multiple choice

The velocity of a body, during a certain time interval, increases from 4 m/s to 6 m/s due to an acceleration of 5 m/s². Find the displacement of the body during this period.

  • A2 m
  • B4 m
  • C1 m
  • D10 m

Answer A. 2 m

  1. We don't know the time, so use the equation without time: v² = v0² + 2as.
  2. 6² = 4² + 2(5)s → 36 = 16 + 10s.
  3. 10s = 20 → s = 2 m.

Watch out: Pick the equation that contains exactly the quantities you have and the one you want.

Free fall and vertical motion

Use g = 9.8 m/s² pointing downward and ignore air resistance unless the question says otherwise.

Q45Multiple choice

For an object in free fall, which of the following is true?

  • AIts acceleration is 9.8 m/s²
  • BIts velocity remains constant
  • CIts acceleration is zero
  • DIts motion depends on its mass

Answer A. Its acceleration is 9.8 m/s²

  1. In free fall the only force is gravity, giving a constant acceleration g = 9.8 m/s² (downward).
  2. Velocity keeps changing (B is wrong), acceleration isn't zero (C is wrong), and mass doesn't matter (D is wrong).
Q46True / False

A free-falling object is an example of motion with constant acceleration.

  • True
  • False

Answer True

  1. Near the Earth's surface, g stays essentially constant at 9.8 m/s².
  2. So free fall is motion with constant acceleration. True.
Q47True / False

Near the Earth's surface, all objects fall with different accelerations depending on their mass.

  • True
  • False

Answer False

  1. Without air resistance every object falls with the same acceleration g = 9.8 m/s².
  2. Heavier objects feel a larger force, but they also have more inertia, so the effects cancel. False.
Q48Multiple choice

A feather and a stone are dropped from the same height at the same time. The feather reaches the ground later than the stone. The reason for this is:

  • AAir resistance has a greater effect on the feather than on the stone.
  • BThe feather has less mass than the stone.
  • CThe feather experiences a smaller gravitational force.
  • DThe acceleration due to gravity is different for different objects.

Answer A. Air resistance has a greater effect on the feather than on the stone.

  1. In a vacuum both would land together, because g is the same for all objects.
  2. In air, drag matters much more for the light, wide feather than for the dense stone.
  3. So the delay is caused by air resistance.

Watch out: Mass and a smaller gravitational force are true statements about the feather, but they are not why it falls slower.

Q49True / False

An object thrown upward will have zero velocity at its maximum height.

  • True
  • False

Answer True

  1. Going up, the velocity decreases. At the top it reaches 0 for an instant, then reverses.
  2. So v = 0 at the maximum height. True.
Q50True / False

At the highest point of its motion, a ball thrown straight up has zero acceleration.

  • True
  • False

Answer False

  1. Velocity is zero at the top, but acceleration is not.
  2. Gravity still acts, so a = 9.8 m/s² downward the whole time. False.

Watch out: Zero velocity does not mean zero acceleration. This is the most common free-fall mistake.

Q51Multiple choice

A ball is thrown vertically upward with an initial velocity of 25.0 m/s. How much time does it take to reach the maximum height?

  • A2.55 s
  • B3.50 s
  • C4.55 s
  • D5.50 s

Answer A. 2.55 s

  1. At the top, v = 0. Use v = v0 − gt.
  2. 0 = 25 − 9.8t → t = 25/9.8 = 2.55 s.
Q52True / False

If a ball is thrown vertically upward with a speed of 20.0 m/s, then the ball needs 1.5 seconds to reach its maximum height.

  • True
  • False

Answer False

  1. At the top v = 0: t = v0/g = 20/9.8 ≈ 2.04 s.
  2. 2.04 s is not 1.5 s, so the statement is False.
Q53Multiple choice

A stone is thrown vertically upward with an initial speed of 22.0 m/s. What is its speed when it reaches a height of 13.0 m?

  • A15.1 m/s
  • B12.1 m/s
  • C34.0 m/s
  • DZero

Answer A. 15.1 m/s

  1. We know the height, not the time, so use v² = v0² − 2gh (gravity opposes the upward motion).
  2. v² = 22² − 2(9.8)(13) = 484 − 254.8 = 229.2.
  3. v = √229.2 ≈ 15.1 m/s.

Watch out: Zero would be the answer only at the maximum height, which here is 484/19.6 ≈ 24.7 m. The stone is still below that.

Q54Multiple choice

If a ball is thrown vertically downward from a height of 120 m, what will its velocity be just before hitting the ground?

  • Av = 48.5 m/s downward
  • Bv = 0
  • Cv = 84.5 m/s downward
  • Dv = 56.0 m/s downward

Answer A. v = 48.5 m/s downward

  1. No initial speed is given, so treat the ball as released from rest: v0 = 0.
  2. v² = v0² + 2gh = 0 + 2(9.8)(120) = 2352.
  3. v = √2352 ≈ 48.5 m/s, downward.

Watch out: When a problem leaves out the starting speed, the intended answer assumes it starts from rest.

Q55Multiple choice

A stone is dropped from rest from the top of a 500 m tower. How far will it have fallen after 3 seconds?

  • A44.1 m
  • B40.2 m
  • C9.8 m
  • D4.9 m

Answer A. 44.1 m

  1. Dropped from rest: v0 = 0, so y = ½gt².
  2. y = ½(9.8)(3)² = 4.9 × 9 = 44.1 m.

Watch out: The 500 m is a distraction. The stone is nowhere near the ground after 3 s.