Title 02.MOTION IN ONE DIMENSIONS.pdf ✅

Objective Physics Volume-I 135 YCT 02. Motion in One Dimensions (a) Distance and Displacement 1. An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is ______ (a) (b) (c) (d) AP EAMCET-20.08.2021, Shift-II SCRA-1998 DCE-2000 AIIMS-2003 Ans. (c) : From Newton's third equation of motion, v 2 – u 2 = 2as v 2 = 2as [∵ u = 0] We know, Parabola equation, x2 = 2ay 2. The relation between time t and distance x is t =ax2 + bx, where a and b are constants. The acceleration is (a) −2 abv2 (b) 2 bv3 (c) −2 av3 (d) 2 av2 MP PET-2012 BCECE-2006 AIEEE-2005 NCERT-1982 Ans. (c) : Let the acceleration be A. ⇒ t = ax2 + bx ...(i) On differentiating the above equation with respect to t. Then, ( ) ( ) ( ) d d d 2 t ax bx dt dt dt = + dx dx 1 2ax b dt dt = + ( ) dx 1 2ax b dt = + ( ) dx 1 v dt 2ax b = = + ...(ii) Again differentiating the equation number (ii) w.r.t. t ( ) ( ) ( ) ( ) 2 2 2 d d 2ax b 1 1. 2ax b d x dt dt dt 2ax b + − + = + ( ) 2 2 2 dx 2a d x dt dt 2ax b − = + dx v dt ∵ = ( ) 2 2 2 d x 2av dt 2ax b − ∴ = + ...(iii) From equation no. (ii) ( ) 1 v 2ax b = + Acceleration 2 2 2 d x 1 2av dt (2ax b) = = − × + Acceleration = –2av × v 2 Acceleration = –2av3 3. 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) 8 m (b) 2 m (c) 4 m (d) 3 m AIEEE-2003, 2004 JIPMER-2001, 2002 AFMC-2000 AIPMT-1998 Ans. (a) : For first case : u = 40 km/h 5 40 m /s 18   = ×     For third equation of motion we have, v 2 = u 2 + 2as ( ) 2 5 0 40 2a 2 18   = × +     2 200 0 4a 18   = +    
Objective Physics Volume-I 136 YCT 2 1 100 a 4 9   = −     .......(i) For second case :- u = 80 km/h 5 u 80 m / s 18   = ×     v 2 = u 2 +2as 2 5 0 80 2 a s 18   = × + × ×     2 200 2as ___(ii) 9   = −    Putting the value of ‘a’ in equation (ii) 2 2 1 100 200 2 s 4 9 9     − × × × = −         2 2 100 100 200 s 4 9 9 9   × × × =     s 2 2 2 = × s = 2 × 2 × 2 Distance, s = 8 m 4. A particle moves along a straight line OX. At a time t (in seconds) the distance x (in meters) 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 WE JEE-2012 JCECE-2007 AFMC-2006 Ans. (c) : Given that, Distance x = 40 + 12t – t3 dx v dt ∴ = ( ) d 3 v 40 12t t dt = + − v = 12 – 3t2 Final velocity of the particle is zero (v = 0) So, 12 – 3t2 = 0 2 12 t 3 = t 4 = t = 2sec. Distance covered by the particle before coming to rest is x = 40 + 12t – t3 = 40 + 12 × 2 – 23 = 64 – 8 = 56 m 5. A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10s is S1 and that covered in the first 20 s is S2, then (a) s2 = 2s1 (b) s2 = 3s1 (c) s2 = 4s1 (d) s2 = s1 KCET-2012 MP PET-2011 AIPMT 2009 Ans. (c): Particle starts motion from rest, u = 0 Let, acceleration = a m/s2 , time t1 = 10 sec velocity after 10 second = v1 m/s So, v1 = u + at1 v1= 0+ a×10 v1= 10 a m/s And 2 2 1 1 v u 2as = + (10a)2 = 0+2as1 2 1 100a s 50a 2a = = Velocity after t2 = 20 second v2 = u+at2 v2 = 0+a×20 v2 = 20 a And 2 2 2 2 v u 2as = + (20a)2 = 0 + 2as2 s2 = 2 1 100a s 50a 2a = = So, 1 2 s 50a s 200a = 1 2 s 1 s 4 = s2 = 4s1 6. A particle located at x = 0, starts moving along the positive x-direction with a velocity v that varies as v = α x where α is dimensionless constant. The displacement of the particle varies with time as (a) t3 (b) t2 (c) t (d) t1/2 AMU-2015 UPSEE-2010 AIEEE-2006 Ans. (b) : Given, v x = α (velocity in x-direction) Differentiating w.r.t time, dx x dt = α x t 0 0 dx dt x = α ∫ ∫ [ ] x t 0 0   2 x t = α   2 2 x t 4 α = Hence, x ∝ t2 2 constant 4   α   =  
Objective Physics Volume-I 137 YCT 7. The displacement of particle is given by 2 1 2 0 a t a t x = a + - 2 3 What is its acceleration? (a) 2 2a 3 (b) 2 2a 3 − (c) 2 a (d) zero UPSEE - 2006 JIPMER-2006 Ans. (b) : Given that, 2 1 2 0 a t a t x a 2 3 = + − 1 2 dx a 2a t v 0 dt 2 3 = = + − 2 dv 2a a dt 3 = = − 2 2a a 3 − ∴ = 8. A person moves 30 m north and then 20 m towards east and finally 30 2 m in south-west direction. The displacement of the person from the origin will be (a) 10 m along north (b) 10 m along south (c) 10 m along west (d) zero TS-EAMCET-11.09.2020, Shift-I J&K CET- 2004 Ans. (c) : The person starts from A- Then, he moves to B and then to C and finally to D. In ∆DEC, CD2 = ED2 + CE2 ED2 = ( 30 2 ) 2 − (30)2 ED2 = 1800 − 900 ED2 = 900 ED = 30m ∴ The displacement of the person from the origin will be AD = DE − AE = 30 − 20 = 10 m The displacement of the person from the origin is 10m in the west. 9. The bus moving with a speed of 42 km/hr is brought to a stop by brakes after 6 m. If the same bus is moving at a speed of 90 km/hr, then the minimum stopping distance is (a) 15.48 m (b) 18.64 m (c) 22.13 m (d) 27. 55 m MHT-CET-2020 J&K-CET-2014 Ans. (d) : Given, u1= 0, s1 = 6m u2 = 0, s2 = ? u1 = 42 × 5 18 = 11.66m/s u2 = 90 × 5 18 = 25 m/s For first case- 2 2 1 1 1 v u 2as = + 0 = (11.66)2 + 2a × 6 a = 11.66 11.66 12 − × = –11.33 m/s2 For second case- 2 2 2 2 2 v u 2as = + 0 = (25)2 + 2(–11.33) × s2 s2 = 625 2 11.33 × =27.5 m 10. The ratio of displacement to distance for a moving particle is (a) Always less than one (b) Always greater than one (c) Always one (d) One or less than one AP EAMCET (17.09.2020) Shift-II JIPMER-2011 Ans. (d) : Displacement is the shortest length between the start and end of a journey, while distance is the actual length of the path travelled. Hence the distance will always be equal to or greater than the displacement. ∴ Distance ≥ Displacement Displacement 1 Distance ≥ So, Displacement isequaloneor less than one Distance 11. A particle starts moving from rest with uniform acceleration. It travels a distance x in first 2 seconds and distance y in the next 2 seconds. Then (a) y = 2x (b) y = 3x (c) y = 4x (d) y = x AP EAMCET-22.09.2020, Shift-I AP EAMCET(Medical)-2014
Objective Physics Volume-I 138 YCT Ans. (b) : Given that, Let x = distance covered in 2 sec x = ( ) 1 1 2 2 .a.t .a 2 2a 2 2 = = And y = distance covered in next 2 sec y = ( ) ( ) 1 1 2 2 .a 4 – .a. 2 2 2 y = 8a – 2a ∴ y = 6a So, x 2a 1 y 6a 3 = = 3x = y ∴ y = 3x 12. 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 COMEDK-2020 [AIPMT 2008] Ans. (b) : Particle has maximum instantaneous velocity at the point at which the slope is maximum. Therefore, max dx v = = maximum slope dt From figure it is obvious that an point C, slope is maximum hence at this point velocity is maximum. 13. 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 CG PET-2021 AIPMT-1994 Ans. (d): At point C, D and F the slope of curve ds dt       is positive, so instantaneous velocity at these points are positive. At point E the slop of curve ds dt       is negative. So instantaneous velocity of the particle at this point is negative. 14. The velocity of a particle is v = v0 + gt + Ft2 . Its position is x= 0 at t = 0, then its displacement after time (t = 1) is (a) 0 v g F + + (b) 0 g F v 2 3 + + (c) 0 g v F 2 + + (d) 0 v 2g 3F + + JEE Main-17.03.2021, Shift-II AIEEE 2007 Ans. (b) : Velocity of particle (v) = v0 + gt +Ft2 dx v dt ∵ = 2 0 dx v gt Ft dt ∵ = + + dx = (v0 + gt+ Ft2 )dt Now integrating the above equation ( ) x 1 2 0 0 0 dx v gt Ft dt = + + ∫ ∫ [ ] 1 2 3 x 0 0 0 gt Ft x v t 2 3   = + +     ∴ 0 g F x v 2 3 = + + 15. The displacement 'x' (in meter) of a particle of mass 'm' (in kg) moving in one dimension under the action of a force, is related to time 't' (in sec) by, t = x + 3 . The displacement of the particle when its velocity is zero, will be (a) 6 m (b) 2 m (c) 4 m (d) 0 m Karnataka CET-2022 COMEDK 2013 CG PET-2010 Ans. (d) : Given that, t x 3 = + x t 3 = − x = t2 + 9 – 6t We know that, dx v dt = v 2t 6 = − As v = 0 t = 3 sec x at t = 3sec x = t2 + 9 – 6t x = (3)2 + 9 – 6 × 3 x = 0