68 free MCQs on Motion in a Straight Line with worked answers and explanations. Equations of motion, projectile motion, relative velocity. Foundation of all mechanics.
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On a velocity-time graph for uniform acceleration, the slope of the line gives the acceleration and the shaded area under the line gives the displacement.
Q1.
A car travels 60 km in 2 hours. What is its average speed?
Answer: B. 30 km/h
Why: Average speed = distance/time = 60/2 = 30 km/h.
Q2.
Which of the following is a vector quantity?
Answer: D. Displacement
Why: Displacement has both magnitude and direction, making it a vector. Speed, distance, and mass are scalars.
Q3.
A ball is dropped from rest and falls freely. Its acceleration is:
Answer: C. 9.8 m/s²
Why: A freely falling body has acceleration equal to g = 9.8 m/s² directed downward.
Q4.
What does the slope of a displacement-time graph represent?
Answer: C. Velocity
Why: Slope = Δdisplacement/Δtime = velocity.
Q5.
A body moving with uniform velocity has acceleration equal to:
Answer: C. 0 m/s²
Why: Uniform velocity means no change in velocity, so acceleration = 0.
Q6.
A train starts from rest and accelerates at 2 m/s² for 10 seconds. Its final velocity is:
Answer: C. 20 m/s
Why: v = u + at = 0 + 2×10 = 20 m/s.
Q7.
Which equation of motion connects velocity, acceleration, and displacement (without time)?
Answer: C. v² = u² + 2as
Why: v² = u² + 2as directly links velocity and displacement without requiring time.
Q8.
The area under a velocity-time graph gives:
Answer: B. Displacement
Why: Area under v-t graph = displacement.
Q9.
Odometer in a vehicle measures:
Answer: C. Distance travelled
Why: Odometer records the total distance travelled by the vehicle along its path.
Q10.
The shortest straight-line distance between the initial and final positions of a body is its:
Answer: A. displacement
Why: Displacement is the straight-line vector from start to finish, unlike the total path distance.
Q11.
The rate of change of displacement with time is:
Answer: A. velocity
Why: Velocity is the rate of change of displacement.
Q12.
The rate of change of velocity with time is:
Answer: A. acceleration
Why: Acceleration is the rate of change of velocity.
Q13.
Which of these is a scalar quantity?
Answer: A. speed
Why: Speed has only magnitude, so it is a scalar; the others are vectors.
Q14.
The SI unit of velocity is:
Answer: A. metre per second
Why: Velocity is displacement per time, so its unit is m/s.
Q15.
The SI unit of acceleration is:
Answer: A. metre per second²
Why: Acceleration is change of velocity per time, so its unit is m/s².
Q16.
A body covering equal distances in equal intervals of time has ___ motion:
Answer: A. uniform
Why: Equal distances in equal times means constant speed - uniform motion.
Q17.
Distance is a ___ quantity:
Answer: A. scalar
Why: Distance has magnitude only, so it is a scalar.
Q18.
Displacement is a ___ quantity:
Answer: A. vector
Why: Displacement has both magnitude and direction, so it is a vector.
Q19.
The slope of a distance–time graph gives the:
Answer: A. speed
Why: The gradient of a distance–time graph is the speed.
Q20.
When a body is at rest, its velocity is:
Answer: A. zero
Why: A body at rest is not moving, so its velocity is zero.
Q21.
A particle starts from rest and moves with acceleration a = 3t m/s². Its velocity at t = 4 s is:
Answer: B. 24 m/s
Why: v = ∫a dt = ∫3t dt = 3t²/2. At t=4: v = 3×16/2 = 24 m/s.
Q22.
A ball is thrown upward with velocity 20 m/s. Maximum height reached is (g=10 m/s²):
Answer: B. 20 m
Why: H = u²/(2g) = 400/(2×10) = 20 m.
Q23.
A car starts from rest and reaches 72 km/h in 10 seconds. Distance covered is:
Answer: B. 100 m
Why: 72 km/h = 20 m/s. s = (u+v)t/2 = (0+20)×10/2 = 100 m.
Q24.
A body travels half its total distance with speed v<sub>1</sub> and half with v<sub>2</sub>. Its average speed is:
Answer: B. 2v<sub>1</sub>v<sub>2</sub>/(v<sub>1</sub>+v<sub>2</sub>)
Why: For equal distances, average speed = harmonic mean = 2v<sub>1</sub>v<sub>2</sub>/(v<sub>1</sub>+v<sub>2</sub>).
Q25.
In a velocity-time graph, what does the area between the graph and the time axis represent?
Answer: C. Distance or displacement
Why: Area under v-t graph equals displacement. If the graph goes below the axis, that area is negative displacement.
Q26.
A body accelerates uniformly from 0 to 60 m/s in 5 s, then decelerates to rest in 5 s. Average velocity is:
Answer: B. 30 m/s
Why: Total distance = ½×60×5 + ½×60×5 = 150+150=300 m. Total time=10 s. Average = 300/10 = 30 m/s.
Q27.
The slope of a velocity–time graph gives the:
Answer: A. acceleration
Why: The gradient of a velocity–time graph is the acceleration.
Q28.
The area under a velocity–time graph gives the:
Answer: A. displacement
Why: The area under a v–t graph equals the displacement.
Q29.
Which equation of motion directly relates v, u, a and t?
Answer: A. v = u + at
Why: v = u + at links final velocity, initial velocity, acceleration and time.
Q30.
For a body in free fall near the Earth, the acceleration is:
Answer: A. g, directed downward
Why: A freely falling body accelerates downward at g ≈ 9.8 m/s².
Q31.
A body starting from rest (u = 0) with acceleration a has, after time t, a velocity of:
Answer: A. at
Why: With u = 0, v = u + at = at.
Q32.
The displacement of a body moving with uniform velocity v for time t is:
Answer: A. vt
Why: For uniform velocity, displacement = velocity × time = vt.
Q33.
A decelerating car has an acceleration that is:
Answer: A. negative
Why: Deceleration means the acceleration opposes the motion, so it is negative.
Q34.
The equation s = ut + ½at² is used to find the:
Answer: A. displacement
Why: This equation gives the displacement s of a uniformly accelerated body.
Q35.
If a body moves with constant velocity, its acceleration is:
Answer: A. zero
Why: Constant velocity means no change in velocity, so acceleration is zero.
Q36.
The velocity–time graph of a uniformly accelerated body is a:
Answer: A. straight sloping line
Why: Uniform acceleration gives a straight line of constant slope on a v–t graph.
Q37.
The velocity–time graph of a body moving at constant velocity is a:
Answer: A. horizontal straight line
Why: Constant velocity means zero slope - a horizontal line.
Q38.
A body thrown vertically upward has, at its highest point, a velocity of:
Answer: A. zero
Why: At the top of the flight the body momentarily stops, so its velocity is zero.
Q39.
Average velocity is the total displacement divided by the total:
Answer: A. time
Why: Average velocity = total displacement / total time.
Q40.
Instantaneous velocity is the velocity of a body at a particular:
Answer: A. instant of time
Why: Instantaneous velocity is the velocity at a specific moment.
Q41.
A particle moves along x-axis with x = 3t² - 4t + 1 (in meters, t in seconds). Velocity at t=2 s is:
Answer: B. 8 m/s
Why: v = dx/dt = 6t - 4. At t=2: v = 12-4 = 8 m/s.
Q42.
Two particles A and B start from origin simultaneously. A moves along x-axis at 3 m/s; B at 4 m/s along y-axis. Distance between them after 5 s:
Answer: C. 25 m
Why: xA=15, yB=20. Distance = √(225+400) = √625 = 25 m.
Q43.
A ball is projected at 60° with initial speed 40 m/s. Height at which it makes 30° with horizontal (g=10):
Answer: C. 50 m
Why: At angle 30°: vx = v·cos30° and vx = u·cos60° = 20. So v = 20/cos30° = 40/√3. vy = v·sin30° = (40/√3)(1/2) = 20/√3. Using v²=u²-2gh for vertical: (20/√3)²=(40sin60°)²-2(10)h → 400/3 = 1200 - 20h → h = (1200-400/3)/20 = (3200/3)/20 = 160/3 ≈ 53.3 m. Closest: 50 m.
Q44.
Acceleration of particle = 3t² - 2t. If initial velocity = 1 m/s, velocity at t = 3 s is:
Answer: D. 22 m/s
Why: v = ∫a dt = t³ - t² + C. At t=0, v=1 so C=1. At t=3: v=27-9+1=19 m/s.
Q45.
A ball thrown horizontally at 20 m/s from height h hits ground at 45° angle. h = (g=10):
Answer: B. 20 m
Why: At ground, vy/vx = tan45° = 1, so vy = vx = 20 m/s. vy = gt → t=2 s. h = ½gt² = ½×10×4 = 20 m.
Q46.
A particle starts with zero velocity and has acceleration a = 2 m/s². The distance covered in nth second is:
Answer: B. 2n-1
Why: s<sub>n</sub> = u + a(2n-1)/2 = 0 + 2(2n-1)/2 = 2n-1 m.
Q47.
A particle at height 40 m is thrown horizontally. Another is thrown upward from ground directly below with speed 10 m/s. They collide after (g=10):
Answer: C. 4 s
Why: Vertical pos: P<sub>1</sub>=40-5t², P<sub>2</sub>=10t-5t². Equal: 40-5t²=10t-5t² → 40=10t → t=4 s.
Q48.
A particle moves in a straight line with velocity v = 2√x m/s (x in m). Acceleration at x=2 m is:
Answer: A. 2 m/s²
Why: a = v(dv/dx). dv/dx = 1/√x. a = 2√x × 1/√x = 2 m/s². Constant acceleration = 2 m/s².
Q49.
A body moving at 20 m/s decelerates uniformly at 4 m/s². It comes to rest after:
Answer: A. 5 s
Why: t = v/a = 20/4 = 5 s.
Q50.
A stone dropped from rest falls for 3 s. Its speed then is (g = 10 m/s²):
Answer: A. 30 m/s
Why: v = gt = 10 × 3 = 30 m/s.
Q51.
A car accelerates from rest at 2 m/s² for 5 s. The distance covered is:
Answer: A. 25 m
Why: s = ½at² = ½ × 2 × 25 = 25 m.
Q52.
The displacement in the nth second, sₙ = u + (a/2)(2n − 1), has units of:
Answer: A. metre
Why: Although "per second" appears in the name, sₙ is a displacement, so its unit is the metre.
Q53.
A body moving at 10 m/s accelerates at 2 m/s² for 3 s. Its final velocity is:
Answer: A. 16 m/s
Why: v = u + at = 10 + 2 × 3 = 16 m/s.
Q54.
A ball thrown vertically up at 20 m/s reaches its highest point after (g = 10 m/s²):
Answer: A. 2 s
Why: t = u/g = 20/10 = 2 s.
Q55.
The maximum height reached by a ball thrown up at 20 m/s is (g = 10 m/s²):
Answer: A. 20 m
Why: H = u²/2g = 400/20 = 20 m.
Q56.
A velocity–time graph is a horizontal line above the time axis. The body has:
Answer: A. constant velocity, zero acceleration
Why: A horizontal v–t line means the velocity is constant, so the acceleration is zero.
Q57.
Using v² = u² + 2as, a body starting from rest reaches 10 m/s over 25 m. Its acceleration is:
Answer: A. 2 m/s²
Why: 100 = 0 + 2a(25), so a = 100/50 = 2 m/s².
Q58.
Two bodies fall freely from different heights in a vacuum. They share the same:
Answer: A. acceleration
Why: In free fall every body has the same acceleration g, regardless of mass or height.
Q59.
The area under an acceleration–time graph gives the change in:
Answer: A. velocity
Why: The area under an a–t graph equals the change in velocity.
Q60.
A body moving with uniform velocity has an acceleration–time graph that is:
Answer: A. along the time axis, at zero
Why: Uniform velocity means zero acceleration, so the a–t graph lies on the time axis.
Q61.
A ball is dropped from rest and travels 25 m in the last second of its fall (g = 10 m/s²). The total height of fall is:
Answer: A. 45 m
Why: Distance in last second = g(t − 0.5) = 25 → t = 3 s; height = (1/2)(10)(9) = 45 m.
Q62.
A stone is thrown vertically up at 30 m/s (g = 10 m/s²). The time interval between the two instants when it is at a height of 40 m is:
Answer: B. 2 s
Why: 40 = 30t − 5t² → t² − 6t + 8 = 0 → t = 2 s and 4 s; interval = 2 s.
Q63.
A particle moves along a line with position x = t³ − 6t² + 9t (metres). Its acceleration is zero at times:
Answer: B. 2 s only
Why: v = 3t² − 12t + 9, a = 6t − 12 = 0 → t = 2 s only.
Q64.
Rain falls vertically. To a man walking at 4 km/h, the rain appears to come at 30° from the vertical. The actual speed of the rain is:
Answer: A. 4√3 km/h
Why: tan30° = 4/v<sub>rain</sub> → v<sub>rain</sub> = 4/tan30° = 4√3 km/h ≈ 6.9 km/h.
Q65.
A ball is dropped from height h and after each bounce rebounds to half the previous height. The total path length travelled before it comes to rest is:
Answer: B. 3h
Why: Total distance = h + 2(h/2 + h/4 + ...) = h + 2·(h/2)/(1 − 1/2) = h + 2h = 3h.
Q66.
A car covers the first half of a distance at 40 km/h and the second half at 60 km/h. Its average speed for the whole trip is:
Answer: B. 48 km/h
Why: For equal distances, average = 2·40·60/(40+60) = 4800/100 = 48 km/h.
Q67.
A particle starts from rest with uniform acceleration. The ratio of the distance covered in the 5th second to that covered in the 1st second is:
Answer: B. 9:1
Why: Distance in nth second ∝ (2n − 1), so ratio = (2·5 − 1):(2·1 − 1) = 9:1.
Q68.
A body moving at 20 m/s decelerates uniformly at 2 m/s². The distance it covers during the 6th second of its motion is:
Answer: A. 9 m
Why: s<sub>6th</sub> = u + a(n − 0.5) = 20 + (−2)(5.5) = 9 m (body stops at t = 10 s, so still moving).