Physics-10.2-Description of motion in one dimension

DESCRIPTION OF MOTION IN ONE DIMENSION SYLLABUS Motion in a straight line, uniform and non-uniform motion, their graphical representation, uniformly accelerated motion, and its applications. 1. 2.1. KINEMATICS AND DYNAMICS The part of mechanics which deals with the description of the motion of an object without considering reason of the origin is called kinematics. Where as, the study of the motion of an object related to its cause is called dynamics. (i) Motion in one dimension: Motion of an object in a straight line is called one dimensional (1-D) motion. The position of a particle in one dimensional motion can be described by only one variable (say x). For a particle moving along a straight line (1-D motion) all the vector quantities such as position, velocity, displacement and acceleration have only one non-zero component. (ii) Motion in two dimension: Motion of an object in a plane is called two dimensional (2-D) motions. For 2-D motion velocity or acceleration can be described by two components in any two mutually perpendicular directions in cartesian coordinate system i.e. its position, velocity, displacement and acceleration can have two nonzero components. (iii) Distance & Displacement: Displacement: The change in position of a body in a particular direction is known as displacement. It is a vector quantity and its unit is meter in SI. The shortest distance between the initial and final positions of the object in a specified direction. Distance: The total length of actual path traversed by a body in a certain interval of time is called distance. It is the actual path travelled by an object between its initial and final positions. It is a scalar quantity and its unit in SI is meter. Displacement may be positive, negative or zero but distance is always positive. If a particle moves in a straight line without change in its direction, the magnitude of displacement is equal to the distance travelled. Otherwise it is always less than it. Thus, Displacement distance Illustration 1: What will be the distance and displacement while moving in a circle from A to B and then B to A as shown in adjoining figure? Key concept: Remember difference between distance and displacement. Solution: Physical quantity and direction Half cycle AB or BA Full cycle AA via B Distance Displacement Direction of displacement R 2R 1. AB, when particle moves from A to B. 2. BA, when particle moves from B to A. 2R 0 (iv) Average Speed and Velocity The average speed of a particle in a given interval of time is defined as the ratio of the distance travelled to the time taken while, average velocity is defined as the ratio of the displacement to the time taken. If a particle moves from A to C through a path ABC. Then distance travelled is the actual path length ABC, while the displacement is, Thus, if the distance travelled is and displacement of a particle is in a given time interval then (v) Instantaneous Speed and Velocity Instantaneous speed and velocity are defined at a particular instant and are given by (vi) Average and Instantaneous Acceleration Average acceleration is defined as the change in velocity over a time interval . Hence, The instantaneous acceleration of a particle is the rate at which its velocity is changing at that instant i.e., Example 2. A particle moves along a semi circle path A to B in a time T as shown in the following fig. (a) Determine the average speed of the particle. (b) Determine the average velocity of the particle. Solution: (a) The average speed of the particle = (b) The average velocity of the particle= 1.1. EQUATIONS OF MOTION Following are the three equations of motion for an object with constant acceleration. (a) (b) (c) where u is the initial velocity of the body (if body start from rest u = 0 ), v is the final velocity, s = displacement travelled by the body in time t seconds and a = acceleration of the body (take + sign for acceleration and – for retardation). The displacement by the body in nth second is given by 1.2. Equation of motion on an inclined plane Let a body of mass m slip down a plane, which is inclined at an angle  with the horizontal. If at t = 0, the body is at top of the inclined plane, then in this case u = 0 and a = g sin (i) In this case the equations of motion are (a) (b) (c) (ii) If time taken by the body to reach the bottom is t, then But or (iii) The velocity of the body at the bottom (iv) The velocity of a body moving on an inclined plane does not depend on the inclination of a plane but the time taken to reach the bottom of the plane depends on the inclination of the plane. The velocity and the time taken by the body on an inclined plane depend on the height. (v) When a body moves on an inclined plane. it traverses one-fourth of the length of the inclined. Plane in time interval 0 to t/2 and the remaining three fourth in the time interval t/2 to t. Example 3. A ball is projected with a velocity of 20 m/s vertically. Find the distance travelled in first three second. (use g = 10m/sec2 ) Solution: Problem here is to find the distance. We can calculate that the direction of ball is changes at t = 2s. (From v = u + at, since v = 0 at highest point therefore 0=20 - 10t  t = 2s) Distance traveled in first two second ( Distance = Displacement, because of velocity does not change direction in one dimension) = 40 – 20 = 20 m (upward) Distance traveled in next one second So total distance travelled by the ball in first three seconds = 20 +5 = 25m Example 4. A particle moves along the x-axis according to x = 4t – t2. Find the distance travelled from t = 0s to t = 3s. Solution: The direction of velocity will change when V = 0 at t = 2 sec. Distance = = = 3+1+ 1 =5 meter Example 5. A body having uniform acceleration of 10 ms-2has a velocity of 100 m s-1. In what time, the velocity will be doubled? Solution: 200 = 100 + 10 t or t = 10 s Example 6. A particle moving with uniform acceleration from A to B along a straight line has velocities v1 and v2 at A and B respectively. If C is the mid point between A and B then determine the velocity of the particle at C. Solution: Let distance between A and B be x and V be the velocity of the particle at C, then V2 = v12 + 2a (x/2) Where a is the acceleration of the particle; i.e. t is the time taken by the particle to travel from A to B and x is the total distance between A and B, i.e. Thus, V2 –v12= Example 7. The velocity of a body travelling with a uniform acceleration of 2 m s-2 is 10 m s-1. What is its velocity after a time interval of 4 s? Solution:  = 10 + 2  4 = 18 m s-1 2. x – t, v – t & a – t GRAPHS FOR MOTION IN ONE DIMENSION (i) Variation of displacement (x), velocity (v) and acceleration (a) with respect to time for different types of motion. Displacement(x) Velocity(v) Acceleration (a) a. At rest b. Motion with constant velocity c. Motion with constant acceleration d. Motion with constant deceleration (ii) Displacement calculation from Velocity - Time Graphs The displacement during an interval between time ti and tf is the area bounded by the velocity curve and the two vertical lines t = ti and t = tf, as shown in figures (a) and (b). (a) For each segment of motion, the velocity is constant. The displacement x1 during the i th interval is the area v1 t1. So the total displacement is x x = v1 t1 + v2 t2 + v3 t3 + v4 t4 (b) When v vs. t graph is a smooth complex curve. Area bounded by the curve and time axis between t=ti and t=tf is the displacement. The area under the curve may be obtained by using integration. (iii) Velocity calculation from Acceleration - time Graphs Given an acceleration–versus–time graph, the change in velocity between t = ti and t = tf is the area bounded by the acceleration curve and the vertical lines t = ti and t = tf (a)The area under the a vs t smooth curve is the change in velocity (b) When a vs t graph is a complex curve, the area under the curve may be obtained by using integration Example 8. The velocity-time graph of a particle moving along a straight line is shown in following figure. (i) If the particle starts its motion from x = –4m, then draw the (a–t) and (x–t) graphs. (ii) Find the displacement of the particle at t = 3 s Key concept: Use uniform acceleration concept. Solution: (i) (ii) 3. Relative Velocity When two objects moves in the same straight line, we compare their motion in terms of their relative velocity. If two objects A and B are moving in a straight line with velocities VA and VB respectively, the relative velocity of object A with respect to object B is given by where VB is called reference object velocity It follows that the relative velocity of object B with respect to object A will be where VA is called reference object velocity Key points regarding relative motion while calculating relative velocity: (i) Relative velocity of a particle = velocity of a particle - velocity of reference object (ii) If the velocity of a particle be VA and the velocity of a reference object be VB then the relative velocity of the particle (iii) Relative velocity of a particle while moving in the same direction. Relative velocity (iv) Relative velocity of a particle while moving in the opposite direction. Relative velocity Example 9 :The position of a particle moving on a straight line path is given by: metre Its velocity at t = 2s is : (A) 84 ms–1 (B) 72 ms–1 (C) 54 ms–1 (D) 36 ms–1 Sol: (C) Velocity = dx/dt = 18 + 18t. It depends upon time. For t = 2s, the velocity = 18 + 18  2 ms–1 = 54 ms–1. OBJECTIVE 1. The position of a particle moving on a straight line path is given by : metre Its acceleration at t = 5 5s is: (A) 9 ms–2 (B) 12 ms–2 (C) 18 ms–2 (D) 45 ms–2 Sol: (C) Acceleration = d2x/dt2 = 18. It does not depend upon time. 2. The motion of a body is given by the equation Where at time t is in ms–1 and t is in seconds. If the body was at rest at t = 0, test the correctness of the following results are. The terminal speed is (A) 2 ms–1 (B) 2 ms–2 (C) 3 ms–1 (D) 2 ms 3. The motion of a body is given by the equation Where at time t is in ms–1 and t is in seconds. If the body was at rest at t = 0, test the correctness of the following results are. The magnitude of the initial acceleration is (A) 5 ms–2 (B) 6 ms–2 (C) 8 ms–2 (D) 7 ms–2 4. The motion of a body is given by the equation Where at time t is in ms–1 and t is in seconds. If the body was at rest at t = 0, test the correctness of the following results are. Then speed is? (A) (B) (C) (D) 5. The motion of a body is given by the equation Where at time t is in ms–1 and t is in seconds. If the body was at rest at t = 0, test the correctness of the following results are. The speed varies is, when the acceleration is half the initial value. (A)  = 2 ms –1 (B)  = 3 ms –1 (C)  = 4 ms –1 (D)  = 1 ms –1 Solution:(2-5) The acceleration of the body at time t is 2. (A) The terminal speed is the speed when the acceleration is zero. Setting a(t) = 0 in Eq.(i), we have 0 = 6 – 3 or terminal speed () = 6/3 = 2 ms–1 3. (B) Putting t = 0 in eq.(1), the initial acceleration is a(0) = 6 – 3(0). Since the body is at t = 0; (0) = 0. Hence a(0) = 6 – 0 = 6 ms–2 4. (C) Eq.(1) can be written as integrating, we have where C is the constant of integration. Now at t = 0. Using this in eq.(2) we have Using this value of C in eq.(ii) we have 5. (D) When we have from eq. (1) 3 = 6 – 3 or  = 1 ms –1 6. A body thrown vertically upward with velocity 10 m/s then how many height traveled by body. (A) 5 m (B) 10 m (C) 15 m (D) 20 m Solution:(A) 7. A driver applies brakes to the vehicle on seeing traffic signal 400 m ahead. At the time of applying the brakes vehicle was moving with 15 ms-1 and then starts retarding with 0.3 ms-2. The distance of vehicle after 1 min from the traffic light: (A) 25m (B) 375m (C) 360m (D) 40m Solution: The maximum distance covered by the vehicle before coming to rest The corresponding time . Therefore after 50 seconds, the distance covered by the vehicle = 375 m from the instant of beginning of braking.  The distance of the vehicle from the traffic after one minute = (400 - 375) m = 25 m  (A) is the correct answer. 8. Which of the following graph correctly represents velocity-time relationship for a particle released from rest to fall freely under gravity? (A) (B) (C) (D) Solution: Releasing of the particle from rest means that v0 =0 at t = 0 and v =gt at any time t.  the slope of v/t graph is a constant.  v/t graph is a straight line passing through the origin.  (A) 9. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as where k is a constant, . At the initial moment the velocity of particle is v0 What time will it take to cover that distance? (A) (B) (C) (D) Solution:(A) and 10. A body starts from rest and acquires a velocity of 4 ms–1 during a displacement of magnitude 4m. Its acceleration is: (A) 0.5 ms–2 (B) 1 ms–2 (C) 2 ms–2 (D) 4 ms–2 Sol: (C) . Hence v = 4 ms–1, v0 = 0 and x = 4m. Thus a = 2 ms–2. 11. At height point velocity of body? (A) Max (B) Min (C) Zero (D) None of these Solution: (C) 12. A body starts from rest and moves with a constant acceleration of 20 ms–2. After 10s the velocity will be: (A) 0.2 ms–1 (B) 2 ms–1 (C) 20 ms–1 (D) 200 ms–1 Sol: (D) . 13. A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The displacement of the stone from the point of release after 10 sec is (A) 490 m (B) 510 m (C) 610 m (D) 725 m Solution: (C)  120  49  610 14. A particle moves along X-axis in such a way that its coordinate X varies with time t according to the equation x  (2 – 5t + 6t2) m. The initial velocity of the particle is (A) – 5 m/s (B) 6 m/s (C) – 3 m/s (D) 3 m/s Solution: (A) x  (2  5t  6t2) mt V  V at t  0 is  5 m/s 15. An athlete completes one round of a circular track or radius R in 40 sec. What will be his displacement at the end of 2 min. 20 sec. (A) Zero (B) 2R (C) 2R (D) 7R Solution: (B) 2 min 20sec  120  20  140 sec So displacement  2R 16. 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 Solution: (B) a down < a up So t down > t up 17. A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is (A) (B) (C) (D) Solution: (B) By conservation of energy 18. A particle moving with constant acceleration covers a distance of 30m in the 3rd second. It covers a distance of 50 M is the 5th second. What is the acceleration of the particle ? (A) 3 ms–2 (B) 5 ms–2 (C) 8 ms–2 (D) 10 ms–2 Sol: (D) Use 19. The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s2. The distance covered by the body in the 5th second of its motion is (A) 25 m (B) 45 m (C) 50 m (D) 85 m Solution: (A) S5  u  a (2t  1)  7   4(10  1)  7  18  25mt. 20. Which of the following four statements is false (A) A body can have zero velocity and still be accelerated (B) A body can have a constant velocity and still have a varying speed (C) A body can have a constant speed and still have a varying velocity (D) The direction of the velocity of a body can change when its acceleration is constant Solution: (B) 21. An aero plane is moving with horizontal velocity u at height h. The speed of a packet dropped from it on the earth’s surface will be (g is acceleration due to gravity) (A) (B) (C) 2gh (D) Solution: (A) 22. The displacement of a particle is given by y  a + bt + ct2 –dt4. The initial velocity and acceleration are respectively (A) b, –4d (B) –b, –2c (C) b, –2c (D) 2c, –4d Solution: (C) y  a  bt  ct2  dt4 V  b  2ct  4dt3 A  2c  12dt2 at t  0, v  b, A  2C 23. A truck and a car are moving with equal velocity. On applying the brakes both will stop after certain distance, then (A) Truck will cover less distance before rest (B) Car will cover less distance before rest (C) Both will cover equal distance (D) None Solution: (B) A  24. The distance traveled by a particle is proportional to the squares of time, then the particle travels with (A) Uniform acceleration (B) Uniform velocity (C) Increasing acceleration (D) Decreasing velocity Solution: (A) S  kt2  v  2kt  a  2k  constant . 25. The graph of displacement v/s time is Its corresponding velocity-time graph will be (A) (B) (C) (D) Solution: (A) S vs. t is parabola So V vs. t must be line and at t = 0 26. The initial velocity of a particle is u (at t = 0) and the acceleration a is given by at. Which of the following relation is valid (A) v = u + a t2 (B) (C) v = u + a t (D) v = u Solution: (D) A  at  27. The initial velocity of the particle is 10 m/sec and its retardation si 2 m/sec2. The distance moved by the particle in 5th second of its motion is (A) 1 m (B) 19 m (C) 50 m (D) 75 m Solution: (C) u  10, a  2 v  0  10 2  t  t  5 28. Pick up the correct statements: (A) Area under a-t graph gives velocity (B) Area under a-t graph gives change in velocity (C) Path of projectile as seen by another projectile is a parabola, (D) A body, whatever be its motion, is always at rest in a frame of reference fixed to the body itself. Solution: (A) 29. A body is moving in a circle at a uniform speed . What is the magnitude of the change in velocity when the radius vector describes an angle  (A) (B) (C) (D) Solution: (D) 30. A bicyclist encounter a series of hills uphill speed is always v1 and down hill speed is always v2. The total distance travelled is , with uphill and downhill portions of equal length. The cyclist's average speed is (A) (B) (C) (D) Solution: (D) For n down and n up hills l  2ns 31. Choose the wrong statement (A) Zero velocity of a particle does not necessarily mean that its acceleration is zero. (B) Zero acceleration of a particle does not necessarily mean that its velocity is zero. (C) If the speed of a particle is constant, its acceleration must be zero. (D) None of these Solution: (D) 32. A particle is moving eastward with a speed of 5 m/s. After 10 seconds, the direction changes towards north, but speed remains same. The average acceleration in this time is (A) zero (B) m/s2 towards N-W (C) m/s2 towards N-E (D) m/s2 towards S-W Solution: (B) (j - i) / 2 so direction is W  N 33. A ballast bag is dropped from a balloon that is 300 m above the ground and rising at 13 m/s. The time before the bag hits the ground is [take g = 10 m/s2] (A) 10 sec (B) 9.8 sec (C) 9.5 sec. (D) 9.15 sec. Solution: (D)  300  13t  ½  10  t2 5t2  13t  300  0 t  34. A stone is thrown vertically upwards with a velocity 30 ms–1. If the acceleration due to gravity is 10 ms–2 , what is the distance travelled by the particle during the first second of its motion ? (A) 10 m (B) 25 m (C) 30 m (D) None of the above. Sol: (B) 35. Displacement (x) of a particle is related to time (t) as x = at + bt2 – ct3 where a, b and are constants of motion. The velocity of the particle when its acceleration is zero is given by (A) (B) (C) (D) Solution: (C) x  at  bt2  ct3, v  a  2bt  3ct2 A  2b  6 ct2 A  0  t  Velocity at t   a  36. A particle moves along a parabolic path y = 9x2 in such a way that the x component of velocity remain constant and has a value . The acceleration of the particle is (A) (B) (C) (D) Solution: (D) 37. The co-ordinate of the particle in x-y plane are given as x  2 + 2t + 4t2 and y  4t + 8t2 The motion of the particle is (A) along a straight line (B) uniformly accelerated (C) along a parabolic path (D) nonuniformly accelerated Solution: (A) x  2  2t  4t2 … (1) y  4t  8t2 … (2) So path  x  2   line 38. A car start from rest then after 100 s is velocity become 100 cm/s then. What is acceleration? (A) 2 (B) 5 (C) 1 (D) 4 Solution: (C) 39. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the position of the particle at t = 2s. (A) 2 m (B) +2 m (C) 3 m (D) 1 m 40. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the displacement of the particle in the time interval from t = 0 to t = 4s. (A) 13 m (B) 12 m (C) 10 m (D) 8 m 41. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the average velocity of the particle in the time interval from t=2s to t = 4s. (A) (B) (C) (D) 42. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the velocity of the particle at t =2s. (A) (B) (C) (D) 43. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the acceleration of the particle at time t = 2 sec. (A) (B) (C) (D) 44. The position of a particle moving along the x-axis is given by x = 3t - 4t2 + t3, where x is in meters and t in seconds. Find the average acceleration of the particle for the time interval t = 2 sec to t = 4 sec. (A) (B) (C) (D) Solution: (39-44) 39. (a) x(t) = 3t  4t2 + t3  x(2) = 3  2  4  (2)2 + (2)3 = 6  4  4 + 8 = 2 m. 40. (b) x(0) = 0 x(4) = 3  4  4  (4)2 + (4)3 = 12m. Displacement = x(4)  x(0) = 12 m. 41. (c) 42 (d)  v(2) = 43. (a) 44. (b) 45. A car starts moving with constant acceleration covers the distance between two points 180 meters apart in 6 seconds. Its speed as it passes the second point is 45 m/s. Find its acceleration. (A) 5m/s2 (B) 8m/s2 (C) 3m/s (D) 2m/s 46. A car starts moving with constant acceleration covers the distance between two points 180 meters apart in 6 seconds. Its speed as it passes the second point is 45 m/s. Find its speed when it was at the first point. (A) 15 m/s2 (B) 15 m/s (C) 10 m/s (D) 5 m/s 47. A car starts moving with constant acceleration covers the distance between two points 180 meters apart in 6 seconds. Its speed as it passes the second point is 45 m/s. Find the distance from the first point when it was at rest. (A) 2 m (B) 22 m2 (C) 22.5 m (D) 0.5 m Solution :(45-47) 45. (a) Let the acceleration of the car be a and its speed at first point be u.  v = u + at  45 = u + at (1) Also 180 = ut + (2) Multiplying equation (1) by t and subtracting from (2), we get 180  45t =   180  45  6 =  a = 5m/s2. 46 (b) From (1) u = 45  5  6 = 15 m/s. 47 (c) Let the distance be S  v2 = u2 + 2aS   S = 22.5 m 48. A body is dropped from the top of the tower and falls freely. The distance covered by it after n seconds is directly proportional to (A) n2 (B) n (C) 2n – 1 (D) 2n2 – 1 Sol: (A) Here 49. In the above question, the distance covered in the nth second is proportional to (A) n2 (B) n (C) 2n – 1 (D) 2n2 – 1 Sol: (C) Here 50. In the question 59, the velocity of the body after n seconds is proportional to (A) n2 (B) n (C) 2n – 1 (D) 2n2 – 1 Sol: (B) Here 51. A body when projected vertically up covers a total distance D. During the time of its flight is t. If there were no gravity, the distance covered by it during the same time is equal to (A) 0 (B) D (C) 2D (D) 4D Solution: The displacement of the body during the time t as it comes back to the point of projection  S = 0  v0t  gt2 = 0  During the same time t, the body moves in absence of gravity through a distance D = v.t, because in absence of gravity g = 0  D = …(1) In presence of gravity the total distance covered is = D = 2H = …(2) (A)  (B)  D = 2D Hence, (C) is correct 52. Velocity each type of physical quantity (A) Vector (B) Scalar (C) Tensor (D) None of these Solution (a) 53 Which is correct ? (A) (B) (C) (D) Solution (B) 54 Accretion is which type of physical quantity (A) Vector (B) Scalar (C) Tensor (D) None of these Solution (a) 55 Which is correct ? (A) (B) (C) (D) Solution (B) 56.: A particle starts from rest with acceleration for some time and after achieving a maximum velocity starts retarding at rate and finally comes to rest. If total time taken is then Determine (a) maximum velocity (b) total distance travelled. (a) (b) (c) (d) Solution: (b) Let the particle accelerate for a time then maximum, velocity Since it retards at a rate and finally comes to rest therefore Or Distance travelled = area under (v -t) graph 57. The displacement of a particle moving in one dimension is given by where is in meter and in second. The displacement, when the velocity is zero is (a) 3 m (b) 1 m (c) 1.8 m (d) None of these Key concept: Differentiate the relation & analyse. Solution: (d) 58. A body is moved along a straight line path by a machine delivering constant power. The distance moved by the body in time is proportional to (a) (b) (c) (d) Key concept: Use formula, power = force velocity Solution: So option (a) is correct. 59. The radius vector of a point depends on time as where and are constant vectors. Find the modulus of velocity and acceleration at any time (a) , (b) , (c) , (d) None of these Solution: (a) Velocity Modulus of velocity vector will be Here c and b are modulus of and and is the angle between and which can be written As Hence, (b) Acceleration Hence, 60. A ball is thrown upwards from the top of a tower 40m high with a velocity of 10m/s. Find the time when it strikes the ground. (a) 2sec. (b) 4sec. (c) 8sec. (d) None of these Key concept: use proper sign rule Solution: (B) s =ut+ ½ at2 s=ut+ ½ at2 - 40 = 10t -½10t2 after solving t= 4 sec and t= -2 sec. Taking positive value t= 4sec.