Motion in One Dimension. A body is moving with velocity 3ms towards East. After s its velocity becomes 4ms towards North. The average acceleration of the body is a) 7ms b) 7ms c) 5ms d) ms. A boy standing at the top of a tower of m height drops a stone. Assming g = ms, the velocity with which it hits the grond is a) ms b) 4ms c) 5ms d) ms 3. Position-time graph for motion with zero acceleration is a) b) c) d) b) c) d) 4. The speed-time graph of a particle moving along a solid crve is shown below. The distance traversed by the particle from t = s to t =3s is a) s b) 4 s c) 3 s d) 5 s

5. A boat is sent across a river with a velocity of 8km/h. If the resltant velocity of boat is km/h, then velocity of the river is a) km/h b) 8 km/h c) 6 km/h d) 4 km/h 6. A train is moving slowly on a straight track with a constant speed of m/s. A passenger in that train starts walking at a steady speed of m/s to the back of the train in the opposite direction of the motion of the train so to an observer standing on the platform directly in front of that passenger, the velocity of the passenger appears to be a) c) 4ms b) ms ms in the opposite direction of the train d) Zero 7. A ball thrown vertically pwards with an initial velocity of.4m/s retrn in s. The total displacement of the ball is a).4 cm b) Zero c) 44.8 m d) 33.6 m 8. From the tap of a tower, a particle is thrown vertically downwards with a velocity of m/s. The ratio of distance covered by it in the 3 rd and nd seconds of its motion is (take g = m / s ) a) 5 : 7 b) 7 : 5 c) 3 : 6 d) 6 : 3 3 9. The position x of a particle varies with time t as x = at bt. The acceleration of the particle will be zero at time t eqal to a) a 3b b) b a c) 3 b. A body starts from rest with a niform acceleration. If its velocity after n seconds is v, then its displacement in the last s is d) c a) v( n n+ ) b) v( n + ) n c) v( n ) n d) v( n n )

. A body A is thrown p vertically from the grond with a velocity v and another H body B is simltaneosly dropped from a height H. They meet at a height, if v is eqal to a) gh b) gh c) gh d) g H. The ratios of the distance traversed in sccessive intervals of time by a body, falling 9 from rest, are a) : 3 : 5 : 7 : 9 : b) : 4 : 6 : 8 : :.. c) : 4 : 7 : : 3 :.. d) None of these 3 3. The displacement of a particle, starting from rest (at t = ) is given by s = 6t t. The time in seconds at which the particle will obtain zero velocity again is a) b) 4 c) 6 d) 8 4. A stone is thrown vertically pwards. When the stone is at a height eqal to half of its maximm height, its speed will be m/s, and then the maximm height attained by the stone is (take g = m / s ) a) 5m b) 5 m c) m d) m 5. Figre () and () show the displacement-time graphs of two particles moving along the x-axis. We can say that a) Both the particles are having a niformly accelerated motion b) Both the particles are having a niformly retarded motion c) Particle () is having on niformly accelerated motion which particle () is having an niformly retarded motion d) Particle () is having a niformly retarded motion while particle () is having an niformly accelerated motion.

8 6. Which of the following can be zero, when a particle is in motion for some time? a) Distance b) Displacement c) Speed d) None of These 7. The distance travelled by a particle starting from rest and moving with an 4 acceleration 3 ms, in the third second is a) 6m b) 4m c) 3 m d) 9 3 m 8. A particle moves in a straight line with a constant acceleration. It changes its velocity from ms to ms while passing throgh a distance 35m in t second. The vale of t is a) b).8 c) d) 9 9. A parachtist after bailing ot falls 5m withot friction. When parachte opens, it decelerates at ms. He reaches the grond with a speed of 3ms. At what height, did he bail ot? a) 9 m b) 8 m c) 93 m d) m. A car, starting from rest, acceleration at the rate f throgh a distance S, then contines at constant speed for time t and then decelerates as the rate f/ to come to rest. If the total distance travelled is 5S, then a) S = ft b) S = ft c) S = ft d) S = ft 6 7 4. A body stats from rest and moves with niform acceleration. Which of the following graphs represent its motion? a) b) c) d)

7. A car moves from X to Y with a niform speed v and retrns to Y with a niform speed v d. The average speed for this rond trip is a) vdv v + v d b) d vdv v v c) v + v d d) v + vd 3 3. The position x of a particle with respect to time t along X-axis is given by x = 9t t where x is in metre and t in second. What will be position of this particle when it achieves maximm speed along the +x direction? a) 3m b) 54m c) 8m d) 4m 4. A particle starting from the origin (, ) moves in a straight line in the (x, y) plane. Its coordinates at a later time are ( 3,3). The path of the particle makes with the X- axis an angle of a) 3 b) 45 c) 6 d) t 5. A particle moving along X-axis has acceleration f, at time t, given by f = f T, where f and T is constant. The particle at t = has zero velocity. In the time interval between t = and the instant when f =, the particle s velocity ( v ) is 3 a) ft b) ft c) ft d) f T 6. A man throws balls with the same speed vertically pwards one after the other at an interval of s. What shold be the speed of the throw so that more than two balls are in the sky at any time? (Given g = 9.8ms ) a) Any speed less than 9.6ms b) Only with speed 9.6ms c) More than 9.6ms d) At least 9.6ms 7. A conveyor belt is moving horizontally at a speed of 4ms. A box of mass kg is gently laid on it. It takes. s for the box to come to rest. If the belt contines to move niformly, then the distance moved by the box on the conveyor belt is a) Zero b). m c).4 m d).8 m x

8. The acceleration of a particle is increasing linearly with time t as bt. The particle 6 starts from the origin with an initial velocity v o. The distance travelled by the particle in time t will be 3 3 a) vot + bt b) vot + bt c) vot + bt d) vot + bt 3 3 6 9. Two spheres of same size, one of mass kg and another of mass 4kg, are dropped simltaneosly from the top of Qtab Minar (height = 7m). When they are m above the grond, the two spheres have the same a) Momentm b) Kinetic Energy c) Potential Energy d) Acceleration 3. The velocity of a particle at an instant is ms. After 3s its velocity will become 6ms. The velocity at s, before the given instant wold have been a) 6 ms b) 4 ms c) ms d) ms 3. A body falls from a height h = m. The ratio of distance travelled in each s, 6 dring t = to t = 6s of the jorney is a) : 4: 9 b) : : 4 c) : 3: 5 d) : : 3 3. Consider the given velocity-time graph It represents the motion of a) A projectile projected vertically pward, from a point

b) An electron in the hydrogen atom c) A bllet fired horizontally from the top of a tower d) An object in the positive direction with decreasing speed 33. A body begins to walk eastward along a street in front of his hose and the graph of 5 his displacement from home is shown in the following figre. His average speed for the whole time interval is eqal to a) 8m min b) 8 6m min c) min 3 m d) m min 34. The displacement x of a particle varies with time t as β are positive constants. The velocity of the particle will = +, where a, b, α and αt βt x ae be a) Go on decreasing with time b) Be independent of α and β c) Drop to zero when α = β d) Go on increasing with time 35. When a ball is thrown p vertically with velocity v, it reaches a maximm height of h. If one wishes to triple the maximm height then the ball shold be thrown with velocity. a) 3v b) 3v c) 9v d) 3 / v

36. If an iron ball and a wooden ball of the same radis are released from a height h in vacm, then time taken by both of them, to reach the grond will be a) Zero b) Uneqal c) Roghly eqal d) Exactly eqal 37. Which of the following velocity-time graphs shows a realistic sitation for a body in motion? a) b) c) d) 4 38. An aeroplane flies 4m de north and then 3m de soth and then flies m pwards, the net displacement is a) Greater than m b) Less than c) 4 m d) 5 m 3 39. A body goes km north and then km de east. The displacement of body from its starting point is a) 3km b) 5.km c).36 km d) km 4. A particle shows distance-time crve as given in this figre. The maximm instantaneos velocity of the particle is arond the point a) B b) C c) D d) A

Key ) c ) a 3) d 4) b 5) c 6) d 7) b 8) b 9) c ) d ) b ) a 3) b 4) d 5) c 6) b 7) c 8) d 9) c ) c ) b ) a 3) b 4) c 5) d 6) c 7) b 8) c 9) d 3) a 3) c 3) a 33) b 34) d 35) a 36) d 37) b 38) a 39) c 4) b Hints. change in velocity Average acceleration = total time a v f = v t i + + = = 3 4 9 6 = 5ms. Given, g = ms and h= m We have v = gh = = 4 = ms = + + = 4 4. Distance (.5 ) (.5 ) 5. Given AB= Velocity of boat = 8 km/h AC= resltant velocity of boat=km/h BC = velocity of river = AC AB = 8

= 6 km/h 6. Relative velocity of the passenger with respect to train = v v = passenger train Relative velocity of the passenger with respect to the observer is zero. 7. Displacement can be defined as the distance between initial and final positions of the ball. Since the ball retrns back to its initial position, the displacement is zero. 8. S ( ) S 3rd nd S S = + 3 = 35m = + ( ) = 5m 3rd = nd 7 5 3 9. x = at bt dx Velocity, v = = at 3bt dt Acceleration, ' d x a = = a bt 6 dt Sbstitting for acceleration given, a - 6bt = v. As v = + na a = n a t = 3b Sn = an and distance travelled in ( n ) second is Sn = a n ( ) So the distance travelled in the last s is = Sn Sn an a n ( )

( ) a n = ( n ) a { n n }{ n n } = + ( ) ( ) ( ) v n = n. Sppose the two bodies A and B meat at time t, at height H from the grond. For body B, =, h = H h = t + gt H Hence, = gt H For body A, = v, h= h = t gt H = v t gt So, from Eqs. (i) and (ii), vt gt = gt v vt = gt t = g Ths, we get H v = g g

H = v g v = gh. Initially = Distance travelled in the nth second is given by g hn = + (n ) Distance travelled in the st second is g g h = + ( ) = Distance travelled in the nd second is g 3g h = + ( ) = Distance travelled in 3 rd second g 5g h 3 = + ( 3 ) = The ratio of distances h : h : h : h : h :... = :3: 5 : 7... 3 4 5 3 3. Given s = 6t t ds Velocity v = = t 3t dt If velocity is zero, then = t 3t t= 4s 4. Let be the initial velocity and h be the maximm height attained by the stone So, v = gh h = h =, v = m / s h () Or = h.(i) Again at height h, v = gh

( v = ) = h = h (ii) So, from eqs. (i) and (ii), we have = h h = m 7. Distance travelled by the particle in nth second is Snth = + a ( n ) Where is initial speed and a is acceleration of the particle Hence, n = 3, = 4 a = ms 3 4 S 3rd = + ( 3 ) 3 = 4 5 6 = 3 m 8. Let and v be the first and final velocities of particle and a and s be the constant acceleration and distance covered by it. From third eqation of motion = + v as = + a Or () () 35 3 a = = ms 35 9 Now sing first eqation of motion, V = + at Or v 9 t = = = = 9s a ( / 9) 9. Parachte bails ot at height H from grond. Velocity at A v = gh = 9.8 5 = 98ms The velocity at grond = 3 v ms (given)

= ms Acceleration (given) v v H h = 98 9 = 4 97 = = 4.75 4 H = 4.75 + h = 4.75 + 5 93m. The velocity-time graph for the given sitation can be drawn as below. Magnitdes of slope of OA = f f And slope of BC = f v = ft = t t = t In graph area of OAD gives Distances, S = ft...(i) Area of rectangle ABED gives distance travelled in time t. S = ( ft ) t

Distance travelled in time t = S3 = f ( t ) Ths, S + S + S3 = 5S S + ( ft ) t + ft = 5S S + ( ft ) t + S = 5S S = ft ( ft) t = S (ii) From eqs (i) and (ii), we have S S t t = 6 ( ft) t = ( ft ) t Fro eq (i), we get S = f ( t ) t S = f = ft 6 7. Average speed dis tan ce travlled = time taken Let t and t be times taken by the car to go from X to Y and then from Y to X respectively. XY XY v + v d Then, t + t = + = XY v vd vvd Total distance travelled =XY+XY=XY Therefore, average speed of the car for this rond trip is

v av XY vvd = or vav = v v + v d + v XY vvd 3. The position x of a particle with respect to time t along X-axis x 9t t 3 =.(i) Differentiating eq (i), with respect to time, we get speed, i.e dx d v = = t t dt dt Or v = 8t 3t 3 (9 ).(ii) d Again differentiating eq. (ii), with respect to time, we get acceleration ie. dx d a = = t t dt dt 3 (9 ) Or a = 8 6t (iii) Now, when speed of particle is maximm, its acceleration is zero, ie A = Ie, 8 6y = Or t = 3s Ptting in eq (i), we obtain position of particle at that time x = = 3 9(3) (3) 9(9) 7 = 8 7 = 54m 4. Draw the sitation as shown. OA represents the path of the particle starting from origin O (, ). Draw a perpendiclar from point A to X-axis. Let path of the particle makes an angle θ with the X-axis, then tanθ = slope of line OA

AB 3 = = = OB 3 3 Or θ = 6 t 5. Acceleration f = f T dv t dv Or f = = f f dt T = dt t Or dv = f dt T.. (i) Integrating eq. (i) on both sides t dv = f dt T f t =. + T v f t C.(ii) Where C is constant of integration. Now, when t =, v = So, from eq. (ii), we get, C = f t v = f t. T t As, f = f T.. (iii) t When, f =, = f T Asm f, so, t = T t = T Sbstitting t = T in eq. (iii) then velocity v f T. T f T = = f T x = f T ft 7. From first eqation of motion

8. v = + at = -4 + a x (.) a = 4ms v s = a (4) s = 4 6 s = s =.m 8 dv bt bt dv dtdt v k dt = = = + At t =, v = v k = v We get Now v = bt + v dx bt dt = + = 3 + + At t =, x = k = = + 6 3 x bt vt 3 bt v x vt k 3. From eqation of motion v = + at 6 = 9 + 3a [here = ms, v = 6ms, t = 3s, a = ms ] And = + x ( = reqired velocity) = 6ms ( t = s) 3. From s = t + gt As the body is falling from rest, = s = gt

Sppose the distance travelled in t = s, t = 4s, t = 6s are s, s4 and s6 respectively Now () s = g = g (4) 8 s4 = g = g (6) 8 s6 = g = g Hence, the distance travelled in first two seconds ( s ) = s s = g i ( s ) = s s = 8g g = 6g m 4 ( s ) = s s = 8g 8g = g f 6 4 Now, the ratio becomes = g : 6 g :g = : 3: 5 33. Distance from to 5s =4 m Distance from 5 to s = m Distance from to 5s = 6 m Distance from 5 to s = m So, net distance =4++6+ = m Total time taken = min. Hence, average speed ce( m) ( ) dis tan = = = 6 m min time min 34. Given = + αt βt x ae be So, velocity dx v = dt αt = α + a e = A + B bβe βt

αt Where A = aα e, B = bβe βt The vale of term A = aα e αt decreases and of term time. As a reslt, velocity goes on increasing with time. B = bβe βt increases with increase in 35. = v hg Or = Or h h h v = 3 h h = 3v or