motion in One Dimension mcq. We covered all the motion in one dimension Class 9ICSEmcq in this post for free so that you can practice well for the exam.
Install our MCQTUBE Android App from the Google Play Store and prepare for any competitive government exams for free.
We also provided the motion in one dimension mcqs with answers pdf Class 11 on our website for your convenience.
A train covers a 100-meter distance and stops when the brakes are applied. If the velocity of the train is just tripled and the same retarding force is applied, then the distance covered by the train is
(A) 200 m
(B) 900 m
(C) 400 m
(D) 600 m
Explanation:
The question asks how the stopping distance changes if the train’s speed triples while the braking force is constant.
Stopping distance depends on the kinetic energy and the work done by the brakes. Kinetic energy is proportional to v2, and the work done by the force is W = F · d.
Initially, the train’s kinetic energy is proportional to v2. Tripling the velocity increases kinetic energy by 9 times, since (3v)2 = 9v2. The brakes perform the same work per meter, so the stopping distance scales with the increase in kinetic energy.
Think of pushing a toy car faster across a table: it travels much farther before stopping because its motion contains more energy that friction must dissipate.
Overall, the stopping distance grows significantly with speed due to the square relationship between kinetic energy and velocity under constant braking force.
Option b – 900 m
A stone is dropped from the balloon and reaches the ground in 20 seconds. Find the height of the balloon if it is rising up with a velocity of 30 m/s. (g = 10 m/s²)
(A) 1200 m
(B) 1400 m
(C) 1000 m
(D) 1800 m
Explanation:
The problem asks to determine the height of a balloon when a stone dropped from it takes 20 seconds to reach the ground, given that the balloon rises at 30 m/s.
The stone’s motion combines the upward velocity of the balloon and downward acceleration due to gravity. Its total displacement depends on the initial upward speed and the time it takes to decelerate and fall.
The stone begins with upward velocity equal to the balloon’s speed. Gravity slows it, stops it, and then accelerates it downward. Using the kinematic equation s = ut + (1/2) g t2, the height can be determined from the known time, initial velocity, and acceleration.
This is analogous to dropping a ball from a moving elevator: the elevator’s motion adds to the ball’s starting speed, affecting how long it takes to reach the ground.
Overall, the balloon’s height depends on both the upward velocity and the gravitational acceleration acting on the stone.
Option b – 1400 m
An elevator car, whose floor-to-ceiling distance is equal to 3 m starts ascending with a constant acceleration of 2 m/s². Four seconds after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is (g = 10 m/s²)
(A) 0.964 s
(B) 3.14 s
(C) 1.414 s
(D) 0.7071 s
Explanation:
The question examines how long it takes for a bolt to hit the floor when released from an accelerating elevator.
The bolt’s motion is relative to the car, which is moving upward with acceleration. Its initial velocity equals the elevator’s instantaneous speed at release. Gravity accelerates it downward, while the elevator moves upward, effectively increasing the relative fall distance.
The total fall time can be analyzed using s = ut + (1/2) g t2, where u is the upward velocity of the bolt at release, and s is the distance to the floor.
This is similar to dropping an object inside a moving elevator: the elevator’s motion affects how quickly the object reaches the floor.
Overall, both the elevator’s acceleration and gravity determine the bolt’s fall time.
Option d – 0.7071 s
A bullet is fired with a speed of 500 m/s in order to hit a target 5 m away. If g = 10 m/s², the gun should be aimed
(A) directly toward the target
(B) 0.06 cm above the target
(C) 0.05 cm above the target
(D) 0.01 cm above the target
Explanation:
This problem deals with compensating for gravitational drop in horizontal projectile motion.
The bullet moves horizontally and vertically simultaneously. Gravity causes vertical displacement during the time taken to reach the target horizontally. Using horizontal velocity v and distance x, the flight time can be found as t = x / v. The vertical drop is y = (1/2) g t2, which determines how much above the target the gun should be aimed.
This is analogous to throwing a ball at a friend across a flat field: you aim slightly higher to account for the downward fall.
Overall, the required aim adjustment depends on horizontal speed, target distance, and gravity.
Option c – 0.05 cm above the target
How much time will a man take to catch a bus if the bus starts moving from rest with an acceleration of 2 m/s² when he is 50 m behind the bus and starts running at 15 m/s to catch the bus?
(A) 4 s
(B) 8 s
(C) 10 s
(D) 5 s
Explanation:
The problem calculates when a running man overtakes an accelerating bus.
The bus starts from rest with acceleration, covering s = (1/2) a t2. The man runs at constant speed, covering s = v t. Setting their displacements equal while accounting for the initial 50 m gap gives a quadratic equation in t.
This is similar to racing a car and a cyclist: the faster one overtakes based on relative speeds and accelerations.
Overall, the time depends on the bus’s acceleration, the man’s constant speed, and their initial separation.
Option d – 5 s
Find the retardation of the particle when its velocity becomes zero if its motion along a straight line is given by x = 6 +8t-4t³
(A) 20 m/s²
(B) 15 m/s²
(C) 18 m/s²
(D) 14 m/s²
Explanation:
The question requires the deceleration of a particle whose motion is described by a cubic displacement equation.
Velocity is obtained by differentiating displacement: v = dx/dt = 8 – 12 t2. Retardation occurs when velocity reaches zero. Acceleration, obtained from a = dv/dt = –24 t, is then evaluated at this instant.
Similar to braking a car: acceleration depends on how velocity changes over time, not just position.
Overall, the particle’s retardation is determined by differentiating velocity at the instant it becomes zero.
Option a – 20 m/s²
A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is
(A) equal to the time of fall.
(B) less than the time of fall.
(C) greater than the time of fall.
(D) twice the time of fall.
Explanation:
The question examines how air resistance affects the rise time of a projectile.
Without air resistance, upward and downward times are equal. Air resistance opposes motion, reducing upward speed faster than gravity alone, so the body takes less time to reach the maximum height.
This is similar to throwing a ball in water: resistance slows it faster, shortening ascent.
Overall, upward motion is shorter than downward motion when resistance is present due to the opposing force.
Option b – less than the time of fall.
Calculate the instantaneous velocity of a particle at 4 s if it describes a motion of x = a + bt² where a = 6 cm and b = 4 cm/s².
(A) 32 cm/s
(B) 46 cm/s
(C) 16 cm/s
(D) 25 cm/s
Explanation:
The problem asks for the instantaneous velocity at a specific time for quadratic motion.
Velocity is the time derivative of displacement: v = dx/dt = 2 b t. Substituting t = 4 s gives the instantaneous speed.
This is like measuring how fast a car accelerates from rest under constant acceleration: velocity grows linearly with time.
Overall, instantaneous velocity depends on the coefficient of t² and the chosen time.
Option a – 32 cm/s
Find the percentage increase in the displacement of a particle during the 5th second compared to that in the 4th second when it moves with constant acceleration along a straight line starting from rest.
(A) 28.57%
(B) 36.42%
(C) 46.43 %
(D) 54.63 %
Explanation:
The question asks for relative displacement in consecutive seconds under uniform acceleration.
Displacement during the nth second is sn = u + (1/2) a (2n – 1), starting from rest (u = 0). Calculating s5 and s4 and finding the percentage difference gives the increase.
Analogy: Like a car accelerating steadily, each successive second covers more distance than the previous one.
Overall, percentage increase depends on uniform acceleration and the formula for nth-second displacement.
Option a – 28.57%
Calculate the velocity of the particle at the end of 4 seconds, if the relation between displacement (x) and time (t) is x = 4t³ – 6t² +9t + 1 where x is in meters and t is in seconds.
(A) 112 m/s
(B) 164 m/s
(C) 256 m/s
(D) 153 m/s
Explanation:
The problem requires instantaneous velocity for cubic displacement.
Velocity is the derivative of displacement: v = dx/dt = 12 t² – 12 t + 9. Substituting t = 4 s gives the particle’s speed at that instant.
This is like monitoring a roller coaster: the speed at any moment depends on how position changes over time.
Overall, instantaneous velocity is obtained by differentiating the displacement function and evaluating at the specified time.
Option d – 153 m/s
Calculate the time taken by a 200 m long train moving with a uniform velocity of 60 km/hr to cross a bridge of length 1000 meters.
(A) 50 s
(B) 2 min 4 s
(C) 1 min 30 s
(D) 1 min 12 s
Explanation:
The problem asks for the time required for a train to cross a bridge.
Time is calculated using t = distance / speed. The total distance to cover is the length of the train plus the bridge. Convert speed to m/s for consistent units.
This is similar to measuring how long it takes for a long truck to fully pass over a bridge.
Overall, total crossing time depends on train length, bridge length, and train speed.
Option d – 1 min 12 s
Calculate the relative velocity of the first car with respect to another car if the first car traveling at 60 km/hr overtakes the second car traveling at 40 km/hr.
(A) 50 km/hr
(B) 15 km/hr
(C) 25 km/hr
(D) 20 km/hr
Explanation:
The question requires the relative speed between two moving cars.
Relative velocity is the difference between their velocities when moving in the same direction: vrel = v₁ – v₂. This determines how fast the first car approaches the second.
Analogy: Like two people walking in the same direction, the faster person catches up at a speed equal to the difference.
Overall, relative velocity is a simple subtraction of speeds in the same direction.
Option d – 20 km/hr
If the speed v of a particle moving along a straight line, when it is at a distance x from a fixed point on the line is given by v² = 90x – 8x², then the magnitude of its H acceleration when it is at a distance 2 meter from the fixed point is
(A) 29 m/s²
(B) 36 m/s²
(C) 21 m/s²
(D) 31 m/s²
Explanation:
The problem asks for acceleration using a position-dependent velocity equation.
Acceleration a = dv/dt. Using chain rule: a = (1/2) d(v²)/dx. Substituting x = 2 m into v² = 90x – 8x² and differentiating gives the acceleration.
This is similar to analyzing a car that speeds up differently depending on its position along a track.
Overall, acceleration depends on the derivative of v² with respect to position.
Option a – 29 m/s²
The frame of reference attached to the Eiffel tower is
(A) always inertial.
(B) always non-inertial.
(C) inertial frame of reference for terrestrial objects.
(D) inertial frame of reference for objects in space.
Explanation:
The question tests understanding of inertial and non-inertial frames.
An inertial frame is one that is at rest or moves with constant velocity. The Eiffel tower is fixed to Earth, which rotates and revolves, making it nearly inertial for most terrestrial calculations. For space objects, other frames may be more suitable.
Analogy: Standing on a slowly moving train can feel almost stationary, similar to the Earth’s surface for daily Physics.
Overall, reference frames depend on relative motion and acceleration.
Option c – inertial frame of reference for terrestrial objects.
From the top of the tower, a stone is released. If it covers 26 m in the last second of its motion, the height of the tower is (g = 10 m/s²)
(A) 38 m
(B) 48 m
(C) 56 m
(D) 22 m
Explanation:
The problem asks for the height of the tower using the distance fallen in the last second.
Distance covered in the nth second is sn = u + (1/2) g (2n – 1), with u = 0 for free fall. Using sn = 26 m, calculate total fall time and then total height H = (1/2) g t².
Analogy: Similar to counting how far a dropped ball moves during its last second of fall to determine total height.
Overall, last-second displacement helps find total distance traveled under gravity.
Option b – 48 m
If the velocity-displacement graph of a particle is linear with positive Y and X-axis intercepts vo and Xo respectively, then the acceleration displacement plot for the same would be
(A) with a positive slope and positive intercept on the acceleration axis.
(B) with a negative slope and negative intercept on the acceleration axis.
(C) with a positive slope and negative intercept on the acceleration axis.
(D) with a positive slope and passing through the origin.
Explanation:
The question examines the relation between acceleration and displacement.
Velocity-displacement graph slope gives acceleration. For v vs x linear, a = dv/dt = v dv/dx. The slope and intercepts determine the nature of the acceleration-displacement plot, which may have a slope and intercept derived from the linear v-x relation.
Analogy: Like linking speedometer readings with distance to find acceleration trends.
Overall, the acceleration-displacement relationship is determined from the v-x linearity.
Option c – with a positive slope and negative intercept on the acceleration axis.
Calculate the average speed for the entire journey of a bus, which covers its onward journey with a constant speed of 40 km/h and its return journey with a constant speed of 60 km/h.
(A) 55 km/h
(B) 48 km/h
(C) 25 km/h
(D) 50 km/h
Explanation:
The question asks for average speed over a round trip with different speeds.
Average speed = total distance / total time. For equal onward and return distances, use harmonic mean: vavg = 2 v₁ v₂ / (v₁ + v₂). This accounts for different travel times at different speeds.
Analogy: Driving to a city slower and returning faster yields a harmonic mean speed.
Overall, round-trip average speed is not the arithmetic mean if speeds differ.
Option b – 48 km/h
A 40 kg block when kept on a 10 m long 30° inclined plane initially at rest, it reaches the bottom of the plane. Calculate the time taken by the block to reach the bottom of the plane. (Take g = 10 m/s²)
(A) 6 s
(B) 4 s
(C) 2 s
(D) 1s
Explanation:
The problem asks for the time for a block to slide down an inclined plane under gravity.
Acceleration along plane: a = g sin θ. Using s = (1/2) a t² with s = 10 m and a = 10 sin 30° = 5 m/s², solve for t. The motion is uniformly accelerated.
Analogy: Like sliding a book down a ramp and timing its descent.
Overall, descent time depends on plane length, slope angle, and gravitational acceleration.
Option c – 2 s
A particle is moving with a velocity of 6 m/s towards the east and its velocity changes to 6 m/s north in 6 s. Find the acceleration.
(A) 2.63 N S-E
(B) 2.63 N N-E
(C) 1.414 N N-W
(D) 1.414 N S-W
Explanation:
The problem asks for acceleration when direction changes but speed remains constant.
Acceleration = change in velocity / time. The velocity change is Vector difference: from east to north, Δv = √(6² + 6²) = 6√2 m/s. Divide by Δt = 6 s for magnitude of acceleration.
Analogy: Like turning a car corner at constant speed: the velocity Vector changes, producing acceleration.
Overall, acceleration depends on how fast the velocity Vector changes in direction.
Option c – 1.414 N N-W
What will be the initial velocity of a ball if a man throws it vertically upward and it rises through 20 m and returns to his hands? (g = 10 m/s²)
(A) 20 m/s
(B) 25 m/s
(C) 10 m/s
(D) 15 m/s
Explanation:
The question asks for the initial speed to reach a specific height under gravity.
Maximum height H = v₀² / (2 g). Using H = 20 m, solve for v₀ = √(2 g H). The motion is symmetric: time to rise equals time to fall.
Analogy: Like tossing a ball to a friend so it reaches a target height and returns.
Overall, initial velocity is determined using energy or kinematic relation for vertical motion.
Option a – 20 m/s
Calculate the time taken by the parrot moving with a speed of 6 m/s towards the south direction to cross the 100 m long train which is moving towards the north direction with a speed of 12 m/s.
(A) 5.55 s
(B) 6.99 s
(C) 4.28 s
(D) 2.96 s
Explanation:
The question asks for the time for a parrot and a train moving in opposite directions to pass each other.
Relative speed = sum of their speeds = 6 + 12 = 18 m/s. Time to cross = train length / relative speed = 100 / 18.
Analogy: Like two people running towards each other on a track; they meet faster than their individual speeds.
Overall, relative motion simplifies the calculation of crossing time.
Option a – 5.55 s
A car moving with a speed of 50 km/hr stops, covering a distance of at least 3 m after applying brakes. If the same car is moving at a speed of 100 km/hr, what is the minimum distance covered by it before coming to halt after applying the brakes?
(A) 12 m
(B) 18 m
(C) 15 m
(D) 9 m
Explanation:
The problem explores how stopping distance varies with speed under constant braking.
Stopping distance ∝ v2. Doubling the speed increases stopping distance by 4 times. Initial distance = 3 m, so new distance scales accordingly.
Analogy: Like sliding a ball: twice the speed, much longer distance to stop.
Overall, stopping distance grows with the square of speed when braking force is unchanged.
Option a – 12 m
Calculate the acceleration of a body at t = 2 s, if its displacement is x = 4t² +9t + 1.
(A) 1 m/s²
(B) 4 m/s²
(C) 2 m/s²
(D) 8 m/s²
Explanation:
The question requires acceleration from a time-dependent displacement function.
Velocity v = dx/dt = 8 t + 9. Acceleration a = dv/dt = 8 m/s². Evaluate at t = 2 s for exact acceleration.
Analogy: Like measuring the instantaneous push on a car given its changing speed.
Overall, acceleration is found by differentiating velocity derived from displacement.
Option d – 8 m/s²
Find the distance traveled in the ninth second of the motion of a ball falling from rest in the gravitational field of the Earth. (g = 10 m/s²)
(A) 45 m
(B) 85 m
(C) 65 m
(D) 75 m
Explanation:
The problem asks for displacement in a specific second of free fall.
Distance in nth second: sn = u + (1/2) g (2n – 1), with u = 0. Substituting n = 9 gives displacement during the ninth second.
Analogy: Counting how far a ball drops in each successive second.
Overall, nth-second formula gives accurate distance for any second in uniformly accelerated motion.
Option b – 85 m
No force is required for an object to be in uniform motion because ( Motion in One Dimension mcq )
The question explores Newton’s first law: uniform motion requires no NETforce.
An object continues in a straight line at constant speed if no external force acts. Acceleration is zero, so constant velocity persists without force.
Analogy: Sliding a puck on ice continues smoothly because friction is negligible.
Overall, uniform motion doesn’t require force; only changes in velocity require force.
Option b – its velocity is constant.
A ball is dropped from the top of a tower of a height of 200 m. Simultaneously another ball was thrown upwards from the bottom of the tower with a speed of 60 m/s. They will cross each other after (g = 10 m/s²)
(A) 5.55 s
(B) 3.33 s
(C) 4.44 s
(D) 6.66 s
Explanation:
The problem involves two bodies moving towards each other: one downward, one upward.
Use relative displacement: let t be crossing time. Distance covered by each adds up to 200 m: sdown + sup = 200. Solve s = ut + (1/2) g t² for each and sum to find t.
Analogy: Like two friends running towards each other from opposite ends of a field; meeting time depends on relative speeds.
Overall, crossing time is found by equating sum of distances to total separation.
Option b – 3.33 s
What is the distance traveled by a particle starting from rest, if it accelerates at 4 m/s² for 5 s and then goes with constant speed for 20 s, and then decelerates at 2 m/s² till it stops?
(A) 750 m
(B) 550 m
(C) 450 m
(D) 650 m
Explanation:
The question requires total distance covered during different motion phases.
Phase 1: s₁ = (1/2) a t² = (1/2)·4·5². Phase 2: s₂ = v t = speed after phase 1 × 20. Phase 3: s₃ = v² / (2 a) = deceleration formula. Total distance = s₁ + s₂ + s₃.
Analogy: Like a car accelerating, cruising, then braking to a stop.
Overall, distance is sum of distances in all motion phases.
Option b – 550 m
A brass ball is tied to a thin wire and swung so as to move uniformly in a horizontal circle. Which of the following statements in this regard is/are true? 1. The ball moves with constant velocity 2. The ball moves with constant speed 3. The ball moves with constant acceleration 4. The magnitude of the acceleration of the ball is constant Select the correct answer using the code given below :
(a) 1 only
(b) 1 and 3
(c) 1,2 and 4
(d) 2 and 4 only
Explanation:
The problem examines motion in a horizontal circle.
Velocity is not constant (direction changes), so statement 1 is false. Speed is constant along the circle. Acceleration exists toward the center (centripetal), and its magnitude is constant.
Analogy: Like a carousel horse moving in a circle at steady speed; its direction changes even if speed doesn’t.
Overall, circular motion has constant speed and constant-magnitude centripetal acceleration.
Option d – 2 and 4 only
Which one of the following statements is not correct? ( motion in one dimension mcq )
(a) If the velocity and acceleration have opposite signs, the object is slowing down
(b) If the velocity is zero at an instant, the acceleration should also be zero at that instant
(c) If the velocity is zero for a time interval; the acceleration is zero at any instant within the time interval
(d) If the position and velocity have opposite signs, the object is moving toward the origin
Explanation:
The question tests understanding of kinematic relationships.
If velocity and acceleration have opposite signs, object slows down. Zero velocity doesn’t imply zero acceleration. Position and velocity relations determine direction of motion. Check each statement for contradictions with Newton’s laws.
Analogy: Observing cars: stopping or moving toward a point depends on velocity and acceleration directions.
Overall, correctness depends on applying Newton’s first and second laws.
Option b – If the velocity is zero at an instant, the acceleration should also be zero at that instant
A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during the sixth second compared to that in the fifth second is about
(a) 11%
(b) 22%
(c) 33%
(d) 44%
Explanation:
The problem asks for percentage increase in displacement for consecutive seconds under uniform acceleration.
Displacement in nth second: sn = u + (1/2) a (2n – 1), with u = 0. Calculate s₆ and s₅, then percentage increase = (s₆ – s₅)/s₅ × 100.
Analogy: Like a car accelerating uniformly; each next second covers more distance.
Overall, percentage increase is determined using the nth-second formula for uniformly accelerated motion.
Option d – 44%
An object is moving with uniform acceleration a. Its initial velocity is u and after time t its velocity is v. The equation of its motion is v=u+at. The velocity (along the y-axis) and time (along the x-axis) graph shall be a straight line.
(a) passing through the origin
(b) with x-intercept u
(c) with y-intercept u
(d) with slope u
Explanation:
The question asks about the shape of a velocity-time graph under uniform acceleration.
For uniform acceleration, v = u + a t is linear in t. The slope of the line is acceleration, and the y-intercept is initial velocity u. The graph is a straight line inclined to the time axis.
Analogy: Like a car accelerating steadily; its speed vs. time graph rises uniformly.
Overall, uniform acceleration produces a straight-line velocity-time graph with slope equal to acceleration.
Option c – with y-intercept u
A boy throws four stones of the same shape, size, and weight at equal speed at different initial angles with the horizontal line. If the angles are 15°, 30°, 45°, and 60°, at which angle the stone will cover the maximum horizontally?
(a) 15⁰
(b) 30°
(c) 45°
(d) 60°
Explanation:
The question is about projectile motion and horizontal range.
Horizontal range R = (v² sin 2θ)/g. Maximum occurs when sin 2θ is maximum = 1, i.e., 2θ = 90° → θ = 45°. Other angles produce smaller ranges.
Analogy: Like throwing a water balloon for distance; throwing at 45° covers the farthest.
Overall, for a given speed, 45° gives the maximum horizontal displacement.
Option c – 45°
If you were to throw a ball vertically upward with an initial velocity of 50 m/s, approximately how long would it take for the ball to return to your hand? Assume air resistance is negligible.
(a) 2.5 s
(b) 5.0 s
(c) 7.5 s
(d) 10 s
Explanation:
The question asks for total time of flight for vertical motion under gravity.
Time to rise tup = v₀ / g. Total time = 2 × tup because ascent and descent times are equal when air resistance is negligible.
Analogy: Tossing a ball straight up; it takes the same time to come down as it took to go up.
Overall, total flight time depends directly on initial velocity and gravitational acceleration.
Option d – 10 s
A bullet is fired vertically up from a 400 m tall tower with a speed of 80 m/s. It is taken as 10 m/s², the time taken by the bullet to reach the ground will be
(a) 8 s
(b) 16 s
(c) 20 s
(d) 24 s
Explanation:
The problem asks for total time to hit the ground from a height with an initial upward velocity.
Use equation of motion: s = v₀ t + (1/2)(-g) t², with s = -400 m (downward). Solve the quadratic equation for t to find the total flight time.
Analogy: Like tossing a ball upwards from a balcony; it first rises, then falls past the balcony to the ground.
Overall, flight time is obtained by solving kinematic equation with upward velocity and initial height.
Option c – 20 s
A Food packet is released from a helicopter rising steadily at the speed of 2 m/sec. After 2 seconds the velocity of the packet is (g = 10 m/sec²)
(a) 22 m/sec
(b) 20 m/sec
(c) 18 m/sec
(d) none of these
Explanation:
The question asks for velocity of a packet dropped from a moving helicopter.
Velocity v = u + a t. Initial upward velocity u = 2 m/s, acceleration a = -g (gravity downward). Substitute t = 2 s to calculate current velocity.
Analogy: Like dropping a ball from a slowly ascending drone; it initially moves up, then slows due to gravity.
Overall, velocity changes linearly under gravity from initial upward motion.
Option c – 18 m/sec
A stone thrown upward with a speed of u from the top of the tower reaches the ground with a velocity of 3u. The height of the tower is ( motion in one dimension mcq )
(a) 3u²/g
(b) 4u²/g
(c) 6u²/g
(d) 9u²/g
Explanation:
The problem asks for tower height using velocities at launch and impact.
Use energy principle: initial kinetic energy + potential energy at top = kinetic energy just before impact. Solve H + (u²)/(2g) = (3u)² / (2g) for height H.
Analogy: Like throwing a ball upwards from a balcony and calculating floor distance knowing impact speed.
Overall, height is determined from energy balance between top and ground.
Option b – 4u²/g
A ball is dropped downwards after 1 sec another ball is dropped downwards from the same point. What is the distance between them after 3 sec?
(a) 25m
(b) 20m
(c) 50m
(d) 9.8m
Explanation:
The question involves two balls dropped successively from rest at 1-second intervals.
Distance fallen by first ball: s₁ = (1/2) g × 3². Distance fallen by second ball: s₂ = (1/2) g × 2². Separation = s₁ – s₂.
Analogy: Like dropping two pebbles into a pond one second apart; the first is always ahead.
Overall, separation depends on squared time differences in free fall.
Option a – 25m
A stone is thrown vertically upwards. When the particle is at a height half of its maximum height, its speed is 10m/sec, then the maximum height attained by the particle is (g=10m/sec²)
(a) 8m
(b) 10m
(c) 15m
(d) 20m
Explanation:
The question asks for maximum height given speed at half the maximum height.
Use energy conservation: at half maximum height, kinetic energy = total energy – potential energy. Solve for H using v² = v₀² – 2 g h, with h = H/2.
Analogy: Like climbing a hill; at midpoint, speed reduces because of gained potential energy.
Overall, maximum height can be calculated from energy or kinematic relations.
Option b – 10m
A body released from the top of a tower falls through half the height of the tower in 2 s. At what time shall the body fall through the height of the tower?
(a) 4 s
(b) 3.26s
(c) 3.48 s
(d) 2.828 s
Explanation:
The problem gives time for first half of height and asks total fall time.
Use equations of motion: s = (1/2) g t². For first half, t₁ = 2 s → s = (1/2) g t₁². For full height H = 2 s² g, solve (1/2) g t² = H for t.
Analogy: Like timing a ball falling through two sections of a tower; full fall time is longer than first half.
Overall, total fall time is derived from first-half timing using uniform acceleration formulas.
Option d – 2.828 s
A ball is thrown vertically upward from the ground with a speed of 25.2 m/s. The ball will reach the highest point of its journey in ( motion in one dimension mcq )
(a) 5.14s
(b) 3.57 s
(c) 2.57 s
(d) 1.29 s
Explanation:
The question asks for time to reach maximum height in vertical motion.
At maximum height, v = 0. Use v = u – g t → t = u / g. With u = 25.2 m/s and g = 9.8–10 m/s², solve for t.
Analogy: Tossing a ball upward; time to peak depends on initial speed and gravity.
Overall, ascent time is proportional to initial velocity over acceleration due to gravity.
Option c – 2.57 s
If the distance S covered by a moving car in rectilinear motion with a speed v in time t is given by S= vt, then the car undergoes
(a) a uniform acceleration
(b) a non-uniform acceleration
(c) a uniform velocity
(d) a non-uniform velocity
Explanation:
The question tests understanding of motion type when displacement depends linearly on time.
S = v t implies constant velocity. There is no change in speed or direction, so acceleration is zero. Motion is uniform.
Analogy: Like a car moving on a straight highway at steady speed; it covers equal distance every second.
Overall, uniform velocity implies no acceleration and straight-line motion.
Option c – a uniform velocity
An automobile traveling with a speed of 60 km/h, can apply brakes to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be
(a) 60m
(b) 40m
(c) 20m
(d) 80m
Explanation:
The question explores how stopping distance depends on speed.
Stopping distance ∝ v2. Doubling speed increases stopping distance by factor of 4. Initial distance = 20 m → new distance = 4 × 20 = 80 m.
Analogy: Like sliding a ball faster; it takes longer to stop because kinetic energy increases with square of speed.
Overall, stopping distance grows quadratically with speed when braking force is constant.
Option d – 80m
A stone thrown vertically upwards with a speed of 5 m/sec attains a height H₁. Another stone thrown upwards from the same point with a speed of 10 m/sec attains a height H₂. The correct relation between H₁ and H₂ is
(a) H₂=4H₁
(b) H₂=3H₁
(c) H₁ =2H₂
(d) H₁ = H₂
Explanation:
The problem asks for comparison of maximum heights for different initial velocities.
Maximum height H = v₀² / (2 g). For H₁, v₀ = 5 → H₁ = 25 / 2g. For H₂, v₀ = 10 → H₂ = 100 / 2g = 4 H₁.
Analogy: Like tossing two balls with different speeds; faster throw reaches higher.
Overall, maximum height depends on square of initial speed in vertical motion.
Option a – H₂=4H₁
We covered all the motion in one dimension mcq above in this post for free so that you can practice well for the exam.
Check out the latest mcq content by visiting our mcqtube website homepage.
My name is Vamshi Krishna and I am from Kamareddy, a district in Telangana. I am a graduate and by profession, I am an android app developer and also interested in blogging.
