Motion — Set 5

Correct Answer: C. [L T^-2]

• **[L T⁻²]** = Acceleration = Δv/Δt = [L T⁻¹]/[T] = [L T⁻²]; there is no mass dimension. • **[L T⁻²]** — velocity is [LT⁻¹]; dividing by time [T] gives [LT⁻²]; this matches m/s². • 💡 Wrong-option analysis: [T⁻¹]: this is the dimension of frequency, not acceleration; [M L T⁻²]: this is the dimension of force (F = ma), which includes mass; [L² T⁻¹]: this is the dimension of kinematic viscosity, not acceleration.

Correct Answer: B. 9.8 m/s^2

• **9.8 m/s²** = The standard value of gravitational acceleration near Earth's surface is g ≈ 9.8 m/s² directed downward. • **g = 9.8 m/s²** — often approximated as 10 m/s² in numerical problems; it varies slightly with latitude and altitude. • 💡 Wrong-option analysis: 1.0 m/s²: approximately g on the Moon's surface; 98 m/s²: ten times too large — a common factor-of-10 error; 0.98 m/s²: ten times too small — g/10.

Correct Answer: B. Total displacement / total time

• **Total displacement / total time** = Average velocity is a vector defined as the total displacement (not distance) divided by the total time interval. • **v_avg = Δx/Δt** — for a round trip the displacement is zero, so average velocity is zero even if average speed is nonzero. • 💡 Wrong-option analysis: Total distance / total time: this is the definition of average speed, not average velocity; Change in speed / time: this is closer to acceleration; Speed × time: this gives distance traveled, not velocity.

Correct Answer: A. Speed is scalar, velocity is vector

• **Speed is scalar, velocity is vector** = Speed has only magnitude (e.g. 60 km/h), while velocity has both magnitude and direction (e.g. 60 km/h northward). • **scalar vs vector** — a car going around a curve at constant speed has changing velocity because direction changes. • 💡 Wrong-option analysis: Speed can be negative, velocity cannot: reversed — speed is always ≥ 0; displacement and velocity can be negative; Speed has direction, velocity does not: completely wrong — speed is the directionless scalar, velocity has direction; Speed equals displacement/time always: wrong — speed = distance/time; displacement/time gives the magnitude of velocity only for straight-line one-way motion.

Correct Answer: D. v = u + at

• **v = u + at** = The first equation of motion states that final velocity equals initial velocity plus acceleration times time. • **v = u + at** — if a is negative (retardation) the formula still holds; a is found from F = ma. • 💡 Wrong-option analysis: v = u − at always: only valid when decelerating; the sign of a handles both cases in v = u + at; v = u/t + a: dimensionally inconsistent — u/t has units m/s² not m/s; v = ut + a: dimensionally inconsistent — ut has units of m, not m/s.

Correct Answer: A. s = ut + (1/2)at^2

• **s = ut + ½at²** = The second equation of motion gives displacement in terms of initial velocity, acceleration, and time. • **s = ut + ½at²** — derived by integrating v = u + at; for u = 0 this reduces to s = ½at². • 💡 Wrong-option analysis: s = (v−u)/t: this yields acceleration (m/s²), not displacement (m); s = v + at: dimensionally inconsistent — v is m/s and at is m/s², cannot be added; s = u + at: this is the formula for final velocity v, not displacement.

Correct Answer: B. v^2 = u^2 + 2as

• **v² = u² + 2as** = The third equation of motion relates velocity, initial velocity, acceleration, and displacement without involving time. • **v² = u² + 2as** — used when time is unknown; rearranging: s = (v²−u²)/(2a) directly gives stopping distance. • 💡 Wrong-option analysis: v = u + 2as: dimensionally wrong — 2as has units m²/s², not m/s; v = u + as: missing factor of 2 and wrong dimensions; v² = u² + as: missing the factor of 2 in the acceleration term.

Correct Answer: A. Zero

• **Zero** = Acceleration = dv/dt; if velocity is constant, dv = 0, so a = 0/dt = 0. • **a = 0 when v = constant** — Newton's first law: a body with constant velocity has zero net force acting on it. • 💡 Wrong-option analysis: Constant nonzero: impossible — nonzero acceleration would change velocity, contradicting 'constant velocity'; Undefined: acceleration is well-defined as zero here; Increasing: increasing acceleration would cause velocity to increase, contradicting the constant-velocity premise.

Correct Answer: B. Velocity (speed in magnitude)

• **Velocity (speed in magnitude)** = The slope of a distance–time graph = Δd/Δt, which is the definition of speed. • **slope = Δd/Δt = speed** — for a displacement–time graph the slope gives velocity including direction. • 💡 Wrong-option analysis: Acceleration: acceleration is the slope of the velocity–time graph, not the distance–time graph; Mass: mass is a property of the body — not derivable from a d–t graph slope; Force: force = ma requires mass and acceleration, not readable from a d–t slope.

Correct Answer: B. Displacement

• **Displacement** = Area under a velocity–time graph = ∫v dt = displacement. • **area = ∫v dt** — this is the fundamental relationship between velocity and displacement in kinematics. • 💡 Wrong-option analysis: Acceleration: acceleration is the slope of the v–t graph; Power: power requires force data, not just a v–t graph; Force: force needs mass and acceleration — not the area under a v–t graph.