Motion in a Straight Line Past Paper PDF

Summary

This document is a collection of physics past paper questions on motion in a straight line. It includes multiple-choice questions and calculations. This is intended for undergraduate students studying physics kinematics.

Full Transcript

# MOTION IN A STRAIGHT LINE ## Topic 1 Distance, Displacement, Speed, Velocity and Acceleration - **01** The position of a particle related to time is given by x = (5t²-4t+5)m. The magnitude of velocity of the particle at t = 2 s will be - (a) 06 ms-1 - (b) 10 ms-1 - (c) 14 ms¹...

# MOTION IN A STRAIGHT LINE ## Topic 1 Distance, Displacement, Speed, Velocity and Acceleration - **01** The position of a particle related to time is given by x = (5t²-4t+5)m. The magnitude of velocity of the particle at t = 2 s will be - (a) 06 ms-1 - (b) 10 ms-1 - (c) 14 ms¹ - (d) 16 ms-1 - **02** A passenger sitting in a train A moving at 90 km/h observes another train B moving in the opposite direction for 8s. If the velocity of the train B is 54 km/h, then length of train B is - (a) 200 m - (b) 320 m - (c) 120 m - (d) 80 m - **03** The distance travelled by an object in time t is given by s = (2.5)t². The instantaneous speed of the object at t=5s will be - (a) 25 ms-1 - (b) 5 ms-1 - (c) 62.5 ms-1 - (d) 12.5 ms-1 - **04** Two trains 'A' and 'B' of length 'l' and '4l' are travelling into a tunnel of length 'L' in parallel tracks from opposite directions with velocities 108 km/h and 72 km/h, respectively. If train 'A' takes 35 s less time than train 'B' to cross the tunnel then, length 'L' of tunnel is (Given, L = 60l) - (a) 1200 m - (b) 1800 m - (c) 2700 m - (d) 900 m - **05** A person travels x distance with velocity v₁ and then x distance with velocity v₂ in the same direction. The average velocity of the person is v, then the relation between v, v₁ and v₂ will be - (a) v = v₁+v₂ / 2 - (b) 1/v = 1/v₁ + 1/v₂ - (c) v = v₁v₂ / v₁+v₂ - (d) 1/v = 1/v₁ + 1/v₂ ## Topic 1 Terms Related to Motion - **01** A vehicle travels half the distance with speed v and the remaining distance with speed 2v. Its average speed is - (a) 2v - (b) 4v/3 - (c) 3v/4 - (d) 3v/2 - **02** A person travelling in a straight line moves with a constant velocity v₁ for certain distance 'x' and with a constant velocity v₂ for next equal distance. The average velocity v is given by the relation - (a) 1/v = 1/v₁ + 1/v₂ - (b) v = v₁+v₂ / 2 - (c) v = v₁v₂ / v₁+v₂ - (d) v = (v₁²v₂²)^1/2 / v₁ + v₂ - **03** Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be - (a) (t₁+t₂)/2 - (b) (t₁*t₂)/(t₂-t₁) - (c) (t₁^2 + t₂^2)^1/2 - (d) (t₁-t₂) - **04** If the velocity of a particle is v = At + Bt², where A and B are constants, then the distance travelled by it between 1 s and 2 s is - (a) A + 7B/3 - (b) 3A/2 + 7B/3 - (c) A/2 + 4B/3 - (d) 3A/2 + 4B/3 - **05** Two cars P and Q start from a point at the same time in a straight line and their positions are represented by Xp(t) = at + bt² and XQ(t) = ft-t². At what time do the cars have the same velocity? - (a) a-f / 1+6 - (b) (a-f) / 2(1+6) - (c) (a+f) / 1+6 - (d) (a+f) / 2(1+6) - **06** A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx-2" where, ẞ and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by - (a) -2nß²x^-2n-1 - (b) -2nß²x^-4n-1 - (c) -2ß²x^-2n+1 - (d) -2nß²e^-4n+1 - **07** The motion of a particle along a straight line is described by equation x = 8+12t-t³ where, x is in metre and t in second. The retardation of the particle when its velocity becomes zero, is - (a) 24 ms-2 - (b) zero - (c) 6 ms-2 - (d) 12 ms-2 - **08** A body is moving with velocity 30 m/s towards East. After 10 s, its velocity becomes 40 m/s towards North. The average acceleration of the body is - (a) 7 m/s² - (b) √7 m/s² - (c) 5 m/s² - (d) 1 m/s² - **09** A particle moves a distance x in time t. According to the equation x = (t+5)^-1. The acceleration of particle is proportional to - (a) (velocity)^3/2 - (b) (distance)^2 - (c) (distance)^-2 - (d) (velocity)^2/3 - **10** A particle moving along X-axis has acceleration f, at time t, given by f = fo(1-t/T) where fo and T are constants. The particle at t = 0 has zero velocity. In the time interval between t = 0 and the instant when f = 0, the particle's velocity (v) is - (a) foT - (b) foT² - (c) foT²/2 - (d) v = foT / 2 - **11** A car moves from X to Y with a uniform speed v, and returns to X with a uniform speed v. The average speed for this round trip is - (a) 2vv / (v+v) - (b) (vv)^1/2 / (v+v) - (c) (v+v) / 2 - (d) (v+v)^1/2 / 2 - **12** The position x of a particle w.r.t. time t along x-axis is given by x = 9t² -t³, where x is in metre and t in sec. What will be the position of this particle when it achieves maximum speed along the + x direction? - (a) 32 m - (b) 54 m - (c) 81 m - (d) 24 m - **13** A particle moves along a straight line OX. At a time t (in second), the distance x (in metre) of the particle from O is given by x = 40 + 12 t – t³. How long would the particle travel before coming to rest? - (a) 24 m - (b) 40 m - (c) 56 m - (d) 16 m - **14** A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is - (a) 10 m/s, 0 - (b) 0, 0 - (c) 0, 10 m/s - (d) 10 m/s, 10 m/s - **15** The displacement x of a particle varies with time t as x = ae^-αt + bBe^βt, where a, b, a and ẞ are positive constants. The velocity of the particle will - (a) decrease with time - (b) be independent of a and ẞ - (c) drop to zero when a = ẞ - (d) increase with time - **16** A particle moves along a straight line such that its displacement at any time t is given by s=3t³ +7t² +14+5.The acceleration of the particle at t = 1 s is - (a) 18 m/s² - (b) 32 m/s² - (c) 29 m/s² - (d) 24 m/s² - **17** The position x of a particle varies with time t, as x = at² - bt³. The acceleration of the particle will zero at time t equals to - (a) zero - (b) a / 3b - (c) 2a / 3b - (d) ab - **18** A car accelerates from rest at a constant rate a for some time, after which it decelerates at a constant rate ẞ and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is - (a) (α² + β²)^1/2 / αβ - (b) (α² - β²)^1/2 / αβ - (c) αβt / (α + β) - (d) αβt / (α - β) - **19** A particle moves along a straight line such that its displacement at any time t is given by s=(t³ -6t² +3t+4) m The velocity when the acceleration is zero, is - (a) 3 ms-1 - (b) -12 ms-1 - (c) 42 ms-1 - (d) -9 ms-1 - **20** A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and the last one-third at 60 km/h. The average speed of the bus is - (a) 9 km/h - (b) 16 km/h - (c) 18 km/h - (d) 48 km/h ## Topic 2 Kinematics Equations of Uniformly and Non-uniformly Accelerated Motion - **23** A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes u / 3. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is - (a) 24 cm - (b) 28 cm - (c) 30 cm - (d) 27 cm - **24** A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s¯¹. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take, g = 10 m s¯²). - (a) 60 m - (b) 64 m - (c) 68 m - (d) 56 m - **25** The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second is - (a) 1:4:9:16 - (b) 1:3:5:7 - (c) 1:1:1:1 - (d) 1:2:3:4 - **26** A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take, g = 10 m/s²) - (a) 20 m/s, 5 m/s² - (b) 20 m/s, 0 - (c) 20√2 m/s, 0 - (d) 20√2 m/s,10 m/s² - **27** A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Lets, be the distance travelled by the block in the interval t = n - 1 to t = n. Then, the ratio is - (a) (2n-1) / (2n) - (b) (2n-1) / (2n+1) - (c) (2n+1) / (2n-1) - (d) (2n) / (2n-1) - **28** A person sitting in the ground floor of a building notices through the window of height 1.5 m, a ball dropped from the roof of the building crosses the window in 0.1 s. What is the velocity of the ball when it is at the topmost point of the window? (g = 10 m/s²) - (a) 15.5 m/s - (b) 14.5 m/s - (c) 4.5 m/s - (d) 20 m/s - **29** A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is (g = 10 m/s²) - (a) 340 m - (b) 320 m - (c) 300 m - (d) 360 m - **30** A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t, if the elevator is at rest and in time t₂ if the elevator is moving uniformly. Then which of the following option is correct? - (a) t₁ <t₂ ort, >t, depending upon whether the lift is going up or down - (b) t₁ <t₂ - (c) t₁ >t₂ - (d) t₁ =t₂ - **31** A toy car with charge a moves on a frictionless horizontal plane surface under the influence of a uniform electric field E. Due to the force qE, its velocity increases from 0 to 6 m/s in one second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively - (a) 1 m/s, 3.5 m/s - (b) 1 m/s, 3 m/s - (c) 2 m/s, 4 m/s - (d) 1.5 m/s, 3 m/s - **32** A stone falls freely under gravity. It covers distances h₁, h₂ and hy in the first 5s, the next 5s and the next 5s respectively. The relation between h₁, h₂ and h₃ is - (a) h₁ =2 h₂ =3h3 - (b) h₁=h₂/3=h₃/5 - (c) h₂ =3 h, and h₃ =3h₂ - (d) h₁ = h₂ = h₃ - **33** The displacement x (in metre) of a particle of mass m (in kg) moving in one dimension under the action of a force, is related to time t (in s) by t = √x +3. The displacement of the particle when its velocity is zero, will be - (a) 4 m - (b) 0 m (zero) - (c) 6 m - (d) 2 m - **34** A particle covers half of its total distance with speed v₁ and the rest half distance with speed v₂. Its average speed during the complete journey is - (a) (v₁+v₂)/2 - (b) (v₁v₂)^1/2 - (c) 2v₁v₂ / (v + v₂) - (d) (v₁^2 + v₂^2 ) ^1/2 / (v₁ + v₂) - **35** A boy standing at the top of a tower of 20 m height drops a stone. Assuming, g = 10 ms², the velocity with which it hits the ground is - (a) 20 m/s - (b) 40 m/s - (c) 5 m/s - (d) 10 m/s - **36** A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 ms²) - (a) 74 ms-1 - (b) 55 ms-1 - (c) 40 ms-1 - (d) 60 ms-1 - **37** A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is s, and that covered in the first 20 s is s₂, then - (a) s2 =2s, - (b) s₂ = 35, - (c) s₂ = 4s, - (d) s2 = s₁ - **38** A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms¯¹ to 20 ms¹ while passing through a distance 135 m in ts. The value of t is - (a) 10 - (b) 1.8 - (c) 12 - (d) 9 - **39** The distance travelled by a particle starting from rest and moving with an acceleration 4/3ms in the third-second is - (a) 6 m - (b) 4 m - (c) (10/3) m - (d) (19/3) m - **40** Two bodies A (of mass 1 kg) and B (of mass 3 kg) are dropped from heights of 16 m and 25 m, respectively. The ratio of the time taken by them to reach the ground is - (a) √(5/4) - (b) (12/5) - (c) (5/12) - (d) (4/5) - **41** A man throws balls with the same speed vertically upwards one after the other at an interval of 2 s. What should be the speed of the throw so that more than two balls are in the sky at any time? (Take g = 9.8 m/s²) - (a) Any speed less than 19.6 m/s - (b) Only with speed 19.6 m/s - (c) More than 19.6 m/s - (d) At least 9.8 m/s - **42** If a ball is thrown vertically upwards with speed u, the distance covered during the last t sec of its ascent is - (a) ut - 1/2*gt^2 - (b) (u+gt)t - (c) ut - (d) gt^2 / 2 - **43** A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is 10 m/s, then the maximum height attained by the stone is (g = 10 m/s²) - (a) 8 m - (b) 10 m - (c) 15 m - (d) 20 m - **44** A car moving with a speed of 40 km/h can be stopped after 2 m by applying brakes. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance? - (a) 2880 m - (b) 2 m - (c) 400 m - (d) 6 m - **45** If a car at rest, accelerates uniformly to a speed of 144 km/h in 20s, it covers a distance of - (a) 2880 m - (b) 1440 m - (c) 400 m - (d) 20 m - **46** If a ball is thrown vertically upwards with a velocity of 40 m/s, then velocity of the ball after 2s will be (g = 10 m/s²) - (a) 15 m/s - (b) 20 m/s - (c) 25 m/s - (d) 28 m/s - **47** A body dropped from a height h with initial velocity zero, strikes the ground with a velocity 3 m/s. Another body of same mass dropped from the same height h with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground is - (a) 5 m/s - (b) 12 m/s - (c) 3 m/s - (d) 4 m/s - **48** The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an initial velocity v. The distance travelled by the particle in time t will be - (a) vot + 1/3*bt² - (b) vot + 1/6bt³ - (c) vot+bt³ - (d) vot + 1/2*bt³ - **49** Three different objects of masses m₁, m, and m₃ are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of - (a) m₁: m2: m3 - (b) m₁: 2m2: 3m3 - (c) 1:1:1 - (d) m₁: m₂: m₃ - **50** The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. How far above the ground is the second drop at that instant? (Take g = 10 m/s²) - (a) 1.25 m - (b) 2.50 m - (c) 3.75 m - (d) 5.00 m ## Topic 3 Graphs Related to Motion - **58** The displacement-time graph of two moving particles make angles of 30° and 45° with the X-axis as shown in the figure. The ratio of their respective velocity is - (a) 1:1 - (b) 1:2 - (c) 1:√3 - (d) √3:1 - **59** A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point - (a) B - (b) C - (c) D - (d) A - **60** The displacement-time graph of moving particle is shown below. The instantaneous velocity of the particle is negative at the point - (a) D - (b) F - (c) C - (d) E - **61** Which of the following curves does not represent motion in one dimension? - (a) graph 1 - (b) graph 2 - (c) graph 3 - (d) graph 4 - **62** Let u be the relative velocity of scooter (s) w.r.t. bus (B), then - (a) us = v + uB - (b) us = v - uB - (c) us = (v * uB)^1/2 - (d) us = v * uB - **63** Letv be the velocity of train, B be the velocity of parrot. Velocity of A w.r.t. B is given by AB = A-B…. Relative velocity of the parrot w.r.t. the train = [10-(-5)] ms = 15 ms-1 Time taken by the parrot to cross the train 150/15 = 10 s - **51** A body is thrown vertically upwards from the ground. It reaches a maximum height of 20 m in 5 s. After what time it will reach the ground from its maximum height position? - (a) 2.5 s - (b) 5 s - (c) 10 s - (d) 25 s - **52** A stone released with zero velocity from the top of a tower, reaches the ground in 4 s. The height of the tower is (g = 10 m/s²) - (a) 20 m - (b) 40 m - (c) 80 m - (d) 160 m - **53** The velocity of train increases uniformly from 20 km/h to 60 km/h in 4 h. The distance travelled by the train during this period is - (a) 160 km - (b) 180 km - (c) 100 km - (d) 120 km - **54** A body starts from rest. what is the ratio of the distance travelled by the body during the 4th and 3rd s? - (a) 7/5 - (b) 7/3 - (c) 5/7 - (d) 3/7 - **55** A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s²) - (a) 60 m - (b) 45 m - (c) 80 m - (d) 50 m - **56** What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey? - (a) 4:5 - (b) 7:9 - (c) 16:25 - (d) 1:1 - **57** A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/h and 40 km/h, respectively. The velocity of the car midway between P and Q is - (a) 33.3 km/h - (b) 20√2 km/h - (c) 25√2 km/h - (d) 0.35 km/h - **64** According to conservation of energy, the kinetic energy of car i.e. 1/2 * mu^2 = work done in stopping the car i.e. F * s where, F is the retarding force and s is the stopping distance. For same retarding force, s₂/s₁ = (80/40)^2 = 4 ⇒ s₂ = 4s₁ = 4×2 = 8 m - **43** Let u be the initial velocity and H be the maximum height attained. At height h = H/2, we have v=v₁ =10 m/s ...From third equation of motion, v^2=u²-2gh Negative sign indicates that velocity and acceleration are in opposite direction ...or (10)² =u²-2gH/2 At height H, v = 0 or 0=u²-2gH ...(i) ...(ii) Subtract Eq. (ii) from Eq. (i), we get (10)2 =2gor H = (10)2/2g or H = 10 m - **65** When an object falls freely under gravity, then its speed depends only on its height of fall and is independent of the mass of the object. As all objects are falling through the same height, therefore their speeds on reaching the ground will be in the ratio of 1:1:1.

Use Quizgecko on...
Browser
Browser