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Unformatted text preview: Preface It gives us immense pleasure to present ‘Solutions To Concepts Of Physics’. This book contains solutions to all the exercise problems from ‘Concepts Of Physics 1 and 2’. The problems have been illustrated in detail with diagrams. You are advised to solve the problems yourself instead of using this book. The book is not written by any of our members and is not meant for sale. - Indrajeet Patil (Admin) Aniket Panse (Mod Head) SOLUTIONS TO CONCEPTS CHAPTER – 1 1. a) Linear momentum b) Frequency : : mv –1 = [MLT ] 1 0 0 –1 = [M L T ] T Force [MLT 2 ] –1 –2 = [ML T ] Area [L2 ] 0 0 –1 a) Angular speed = /t = [M L T ] c) Pressure : 2. M0L0 T 2 [M0L0T–2] t T –2 2 –2 c) Torque = F r = [MLT ] [L] = [ML T ] 2 2 2 0 d) Moment of inertia = Mr = [M] [L ] = [ML T ] 2 MLT a) Electric field E = F/q = [MLT 3I1 ] [IT ] b) Angular acceleration = 3. b) Magnetic field B = F MLT 2 [MT 2I1 ] qv [IT ][LT 1 ] B 2a MT 2I1 ] [L] [MLT 2I2 ] I [I] a) Electric dipole moment P = qI = [IT] × [L] = [LTI] 2 2 b) Magnetic dipole moment M = IA = [I] [L ] [L I] E = h where E = energy and = frequency. c) Magnetic permeability 0 = 4. 5. h= 6. E [ML2 T 2 ] [ML2 T 1 ] 1 [T ] Q [ML2 T 2 ] [L2 T 2K 1 ] mT [M][K ] L L2 [L] b) Coefficient of linear expansion = = 1 [K 1 ] L 0 T [L ][R] a) Specific heat capacity = C = PV [ML1T 2 ][L3 ] [ML2 T 2K 1(mol) 1 ] nT [(mol )][K ] Taking force, length and time as fundamental quantity m ( force/acce leration) [F / LT 2 ] F 4 2 [FL 4 T 2 ] a) Density = 2 V Volume [L ] LT 2 –2 b) Pressure = F/A = F/L = [FL ] –2 –1 c) Momentum = mv (Force / acceleration) × Velocity = [F / LT ] × [LT ] = [FT] 1 Force d) Energy = mv 2 ( velocity )2 2 accelerati on F F = 2 [LT 1 ]2 2 [L2 T 2 ] [FL] LT LT ] metre 5 2 g = 10 = 36 10 cm/min 2 sec The average speed of a snail is 0.02 mile/hr 0.02 1.6 1000 –1 Converting to S.I. units, m/sec [1 mile = 1.6 km = 1600 m] = 0.0089 ms 3600 c) Gas constant = R = 7. 8. 9. The average speed of leopard = 70 miles/hr In SI units = 70 miles/hour = 70 1.6 1000 = 31 m/s 3600 1.1 Chapter-I 3 10. Height h = 75 cm, Density of mercury = 13600 kg/m , g = 9.8 ms 4 –2 then 2 Pressure = hfg = 10 10 N/m (approximately) 5 In C.G.S. Units, P = 10 × 10 dyne/cm 2 11. In S.I. unit 100 watt = 100 Joule/sec 9 In C.G.S. Unit = 10 erg/sec 12. 1 micro century = 104 × 100 years = 10–4 365 24 60 min 5 So, 100 min = 10 / 52560 = 1.9 microcentury 13. Surface tension of water = 72 dyne/cm In S.I. Unit, 72 dyne/cm = 0.072 N/m 14. K = kIa b where k = Kinetic energy of rotating body and k = dimensionless constant Dimensions of left side are, 2 –2 K = [ML T ] Dimensions of right side are, Ia = [ML2]a, b = [T–1]b According to principle of homogeneity of dimension, 2 –2 2 –2 –1 b [ML T ] = [ML T ] [T ] Equating the dimension of both sides, 2 = 2a and –2 = –b a = 1 and b = 2 a b 15. Let energy E M C where M = Mass, C = speed of light a b E = KM C (K = proportionality constant) Dimension of left side 2 –2 E = [ML T ] Dimension of right side a a b –1 b M = [M] , [C] = [LT ] 2 –2 a –1 b [ML T ] = [M] [LT ] a = 1; b = 2 So, the relation is E = KMC 2 2 –3 –2 16. Dimensional formulae of R = [ML T I ] 2 3 –1 Dimensional formulae of V = [ML T I ] Dimensional formulae of I = [I] 2 3 –1 2 –3 –2 [ML T I ] = [ML T I ] [I] V = IR a b c 17. Frequency f = KL F M M = Mass/unit length, L = length, F = tension (force) –1 Dimension of f = [T ] Dimension of right side, a a b –2 b c –1 c L = [L ], F = [MLT ] , M = [ML ] –1 a –2 b –1 c [T ] = K[L] [MLT ] [ML ] 0 0 –1 M L T = KM b+c a+b–c L –2b T Equating the dimensions of both sides, b+c=0 …(1) –c + a + b = 0 …(2) –2b = –1 …(3) Solving the equations we get, a = –1, b = 1/2 and c = –1/2 –1 1/2 So, frequency f = KL F M –1/2 = K 1/ 2 1/ 2 K F F M L L M 1.2 Chapter-I 18. a) h = 2SCos rg LHS = [L] MLT 2 [MT 2 ] L Surface tension = S = F/I = –3 0 Density = = M/V = [ML T ] –2 Radius = r = [L], g = [LT ] RHS = 2Scos [MT 2 ] [M0L1T0 ] [L] 3 0 rg [ML T ][L][LT 2 ] LHS = RHS So, the relation is correct b) v = p where v = velocity –1 LHS = Dimension of v = [LT ] –1 –2 Dimension of p = F/A = [ML T ] –3 Dimension of = m/V = [ML ] p [ML1T 2 ] [L2T 2 ]1/ 2 = [LT 1 ] [ML3 ] RHS = So, the relation is correct. 4 c) V = (pr t) / (8l) 3 LHS = Dimension of V = [L ] –1 –2 4 4 Dimension of p = [ML T ], r = [L ], t = [T] –1 –1 Coefficient of viscosity = [ML T ] RHS = pr 4 t [ML1T 2 ][L4 ][T] 8 l [ML1T 1 ][L] So, the relation is correct. d) v = 1 (mgl / I) 2 –1 LHS = dimension of v = [T ] RHS = [M][LT 2 ][L] (mgl / I) = 2 [ML ] –1 = [T ] LHS = RHS So, the relation is correct. 19. Dimension of the left side = Dimension of the right side = So, the dimension of dx 2 (a x ) L 2 2 (a x ) ≠ 0 2 (L L ) 1 1 a –1 sin = [L ] a x dx 2 2 1 1 a sin a x So, the equation is dimensionally incorrect. 1.3 = [L ] Chapter-I 20. Important Dimensions and Units : Physical quantity Force (F) Work (W) Power (P) Gravitational constant (G) Angular velocity () Angular momentum (L) Moment of inertia (I) Torque () Young’s modulus (Y) Surface Tension (S) Coefficient of viscosity () Pressure (p) Intensity of wave (I) Specific heat capacity (c) Stefan’s constant () Thermal conductivity (k) Current density (j) Electrical conductivity () Electric dipole moment (p) Electric field (E) Electrical potential (V) Electric flux () Capacitance (C) Permittivity () Permeability () Magnetic dipole moment (M) Magnetic flux () Magnetic field (B) Inductance (L) Resistance (R) Dimension [M1L1T 2 ] SI unit newton 1 2 2 [M L T ] joule 1 2 3 [M L T ] watt 1 3 2 [M L T ] 2 N-m /kg [T 1] 2 radian/s 1 2 1 [M L T ] 2 kg-m /s [M1L2 ] kg-m 1 2 2 [M L T ] N-m [M1L1T 2 ] N/m [M1T 2 ] 2 N/m [M1L1T 1] N-s/m [M1L1T 2 ] 2 2 N/m (Pascal) [M1T 3 ] watt/m [L2T 2K 1] 1 3 2 2 J/kg-K 4 [M T K ] 2 watt/m -k [M1L1T 3K 1] 4 watt/m-K 1 2 [I L ] ampere/m [I2T3M1L3 ] [L1I1T1] –1 2 –1 m C-m 1 1 1 3 [M L I T ] V/m 1 2 1 3 [M L I T ] volt 1 3 1 3 [M T I L ] 2 4 1 2 2 4 1 3 volt/m [I T M L ] farad (F) [I T M L ] 2 C /N-m 1 1 2 3 [M L I T ] 2 2 Newton/A [I1L2 ] N-m/T [M1L2I1T 2 ] Weber (Wb) 1 1 2 [M I T ] tesla 1 2 2 2 [M L I T ] henry 1 2 2 3 [M L I T ] ohm () **** 1.4 SOLUTIONS TO CONCEPTS CHAPTER – 2 1. As shown in the figure, The angle between A and B = 110° – 20° = 90° | A | = 3 and | B | = 4m B R y 20 A x A 2 B 2 2AB cos = 5 m Let be the angle between R and A Resultant R = 4 sin 90 –1 = tan 1 = tan (4/3) = 53° 3 4 cos 90 Resultant vector makes angle (53° + 20°) = 73° with x-axis. 2. Angle between A and B is = 60° – 30° =30° | A | and | B | = 10 unit B y 60° A 102 10 2 2.10.10.cos30 = 19.3 be the angle between R and A 10 sin30 1 –1 –1 1 = tan tan = tan (0.26795) = 15° 10 10 cos30 2 3 30° R= x Resultant makes 15° + 30° = 45° angle with x-axis. 3. x component of A = 100 cos 45° = 100 / 2 unit x component of B = 100 cos 135° = 100 / 2 x component of C = 100 cos 315° = 100 / 2 Resultant x component = 100 / 2 – 100 / 2 + 100 / 2 = 100 / 2 y component of A = 100 sin 45° = 100 / 2 unit y component of B = 100 sin 135° = 100 / 2 y component of C = 100 sin 315° = – 100 / 2 Resultant y component = 100 / 2 + 100 / 2 – 100 / 2 = 100 / 2 Resultant = 100 y component Tan = =1 x component –1 = tan (1) = 45° The resultant is 100 unit at 45° with x-axis. 4. a 4i 3j , b 3i 4 j a) | a | 42 3 2 = 5 b) | b | 9 16 = 5 c) | a b || 7 i 7 j | 7 2 d) a b ( 3 4)iˆ ( 4 3)ˆj ˆi ˆj | a b | 12 ( 1)2 2 . 2.1 45° 315° 135° Chapter-2 5. x component of OA = 2cos30° = 3 A 2m 30° y x component of BC = 1.5 cos 120° = –0.75 1.5m 60° x 90° D B 1m O x component of DE = 1 cos 270° = 0 y component of OA = 2 sin 30° = 1 E y component of BC = 1.5 sin 120° = 1.3 y component of DE = 1 sin 270° = –1 Rx = x component of resultant = 3 0.75 0 = 0.98 m Ry = resultant y component = 1 + 1.3 – 1 = 1.3 m So, R = Resultant = 1.6 m If it makes and angle with positive x-axis y component = 1.32 Tan = x component –1 = tan 1.32 6. | a | = 3m | b | = 4 a) If R = 1 unit 3 2 42 2.3.4. cos = 1 = 180° 3 2 4 2 2.3.4. cos = 5 b) = 90° 3 2 4 2 2.3.4. cos = 7 c) = 0° Angle between them is 0°. 7. AD 2ˆi 0.5Jˆ 4Kˆ = 6iˆ 0.5 ˆj AD = Tan = DE / AE = 1/12 = tan A 2m (1/12) 0.5 km E B 6m The displacement of the car is 6.02 km along the distance tan 8. D 0.5 km AE 2 DE2 = 6.02 KM –1 4m C –1 (1/12) with positive x-axis. In ABC, tan = x/2 and in DCE, tan = (2 – x)/4 tan = (x/2) = (2 – x)/4 = 4x 4 – 2x = 4x 6x = 4 x = 2/3 ft a) In ABC, AC = AB2 BC2 = C 2 10 ft 3 b) In CDE, DE = 1 – (2/3) = 4/3 ft CD = 4 ft. So, CE = CD2 DE2 = x F BC = 2 ft AF = 2 ft DE = 2x 2–x E G D A 4 10 ft 3 AG2 GE 2 = 2 2 ft. Here the displacement vector r 7ˆi 4 ˆj 3kˆ B c) In AGE, AE = 9. a) magnitude of displacement = z 74 ft b) the components of the displacement vector are 7 ft, 4 ft and 3 ft. 2.2 r Y Chapter-2 10. a is a vector of magnitude 4.5 unit due north. a) 3| a | = 3 4.5 = 13.5 3 a is along north having magnitude 13.5 units. b) –4| a | = –4 1.5 = –6 unit –4 a is a vector of magnitude 6 unit due south. 11. | a | = 2 m, | b | = 3 m angle between them = 60° 2 a) a b | a | | b | cos 60 = 2 3 1/2 = 3 m 2 b) | a b || a | | b | sin 60 = 2 3 3 / 2 = 3 3 m . 12. We know that according to polygon law of vector addition, the resultant of these six vectors is zero. A4 A5 Here A = B = C = D = E = F (magnitude) So, Rx = A cos + A cos /3 + A cos 2/3 + A cos 3/3 + A cos 4/4 + A cos 5/5 = 0 A6 A3 [As resultant is zero. X component of resultant Rx = 0] = cos + cos /3 + cos 2/3 + cos 3/3 + cos 4/3 + cos 5/3 = 0 60° = /3 A1 A2 Note : Similarly it can be proved that, sin + sin /3 + sin 2/3 + sin 3/3 + sin 4/3 + sin 5/3 = 0 13. a 2 i 3 j 4k; b 3 i 4 j 5k 1 a b a b ab cos cos ab 2 3 3 4 4 5 38 cos1 cos 1 2 2 2 2 2 2 1450 2 3 4 3 4 5 14. A ( A B) 0 (claim) As, A B AB sin nˆ AB sin nˆ is a vector which is perpendicular to the plane containing A and B , this implies that it is also perpendicular to A . As dot product of two perpendicular vector is zero. Thus A ( A B) 0 . 15. A 2iˆ 3ˆj 4kˆ , B 4iˆ 3 ˆj 2kˆ ˆi ˆj kˆ ˆ 12) 6iˆ 12ˆj 6kˆ . A B 2 3 4 ˆi(6 12) ˆj(4 16) k(6 4 3 2 16. Given that A , B and C are mutually perpendicular A × B is a vector which direction is perpendicular to the plane containing A and B . Also C is perpendicular to A and B Angle between C and A × B is 0° or 180° (fig.1) So, C × ( A × B ) = 0 The converse is not true. For example, if two of the vector are parallel, (fig.2), then also C × (A × B) = 0 So, they need not be mutually perpendicular. 2.3 B C ( A B) A B C A Chapter-2 17. The particle moves on the straight line PP’ at speed v. From the figure, P Q V P OP v (OP)v sin nˆ = v(OP) sin nˆ = v(OQ) nˆ It can be seen from the figure, OQ = OP sin = OP’ sin ’ So, whatever may be the position of the particle, the magnitude and direction of OP v remain constant. OP v is independent of the position P. 18. Give F qE q( v B) 0 E ( v B) So, the direction of v B should be opposite to the direction of E . Hence, v should be in the positive yz-plane. O y B E Again, E = vB sin v = B sin x E V For v to be minimum, = 90° and so vmin = F/B So, the particle must be projected at a minimum speed of E/B along +ve z-axis ( = 90°) as shown in the figure, so that the force is zero. 19. For example, as shown in the figure, A B B along west BC A along south C along north A B = 0 A B B C B C = 0 But B C C B B A 2 20. The graph y = 2x should be drawn by the student on a graph paper for exact results. To find slope at any point, draw a tangent at the point and extend the line to meet x-axis. Then find tan as shown in the figure. It can be checked that, Slope = tan = y=2x2 y x dy d (2x 2 ) = 4x dx dx Where x = the x-coordinate of the point where the slope is to be measured. y 21. y = sinx So, y + y = sin (x + x) y = sin (x + x) – sin x y = sinx = sin = 0.0157. 3 3 100 22. Given that, i = i0 e t / RC Rate of change of current = When a) t = 0, i di d d i0 e i / RC i0 e t / RC = 0 e t / RC dt dt dt RC di i dt RC i di dt RCe i0 di c) when t = 10 RC, dt RCe10 b) when t = RC, 2.4 x Chapter-2 23. Equation i = i0 e t / RC –5 i0 = 2A, R = 6 10 a) i = 2 e b) 0.3 603 510 7 , C = 0.0500 10 2e 0.3 0.3 –6 –7 F = 5 10 F 2 amp . e di 2 ( 0.3 / 0.3) 20 di i0 t / RC when t = 0.3 sec e Amp / sec e dt 0.30 3e dt RC c) At t = 0.31 sec, i = 2e( 0.3 / 0.3) 5.8 Amp . 3e 2 24. y = 3x + 6x + 7 Area bounded by the curve, x axis with coordinates with x = 5 and x = 10 is given by, y Area = 0 2 sin xdx [cos x]0 dy = y = 3x2 + 6x + 7 10 x3 x2 10 dy = (3x 6x 7)dx = 3 5 7x 5 = 1135 sq.units. 3 5 3 5 5 y 25. Area = 10 10 y 5 10 x =2 0 0 y y = sinx x y –x 26. The given function is y = e When x = 0, y = e –0 =1 x increases, y value deceases and only at x = , y = 0. So, the required area can be found out by integrating the function from 0 to . So, Area = e x x dx [e x ]0 1 . 0 mass 27. a bx length y 2 a) S.I. unit of ‘a’ = kg/m and SI unit of ‘b’ = kg/m (from principle of homogeneity of dimensions) O b) Let us consider a small element of length ‘dx’ at a distance x from the origin as shown in the figure. dm = mass of the element = dx = (a + bx) dx So, mass of the rod = m = 28. L L bx 2 bL2 dm (a bx )dx = ax aL 2 0 2 0 dp = (10 N) + (2 N/S)t dt momentum is zero at t = 0 momentum at t = 10 sec will be dp = [(10 N) + 2Ns t]dt p 10 10 0 0 0 dp 10dt 10 (2tdt) = 10t 0 2 10 t2 = 200 kg m/s. 2 0 2.5 x x =1 Chapter-2 29. The change in a function of y and the independent variable x are related as dy x2 . dx 2 dy = x dx Taking integration of both sides, dy x 2 dx y = x3 c 3 y as a function of x is represented by y = x3 c. 3 30. The number significant digits a) 1001 No.of significant digits = 4 b) 100.1 No.of significant digits = 4 c) 100.10 No.of significant digits = 5 d) 0.001001 No.of significant digits = 4 31. The metre scale is graduated at every millimeter. 1 m = 100 mm The minimum no.of significant digit may be 1 (e.g. for measurements like 5 mm, 7 mm etc) and the maximum no.of significant digits may be 4 (e.g.1000 mm) So, the no.of significant digits may be 1, 2, 3 or 4. 32. a) In the value 3472, after the digit 4, 7 is present. Its value is greater than 5. So, the next two digits are neglected and the value of 4 is increased by 1. value becomes 3500 b) value = 84 c) 2.6 d) value is 28. 33. Given that, for the cylinder Length = l = 4.54 cm, radius = r = 1.75 cm 2 Volume = r l = (4.54) (1.75) r 2 Since, the minimum no.of significant digits on a particular term is 3, the result should have 3 significant digits and others rounded off. 2 l 3 So, volume V = r l = (3.14) (1.75) (1.75) (4.54) = 43.6577 cm 3 Since, it is to be rounded off to 3 significant digits, V = 43.7 cm . 34. We know that, Average thickness = 2.17 2.17 2.18 = 2.1733 mm 3 Rounding off to 3 significant digits, average thickness = 2.17 mm. 35. As shown in the figure, Actual effective length = (90.0 + 2.13) cm But, in the measurement 90.0 cm, the no. of significant digits is only 2. So, the addition must be done by considering only 2 significant digits of each measurement. So, effective length = 90.0 + 2.1 = 92.1 cm. **** 2.6 90cm 2.13cm SOLUTIONS TO CONCEPTS CHAPTER – 3 1. 2. 3. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD W AD = AF2 DF2 302 402 50m In AED tan = DE/AE = 30/40 = 3/4 = tan–1 (3/4) His displacement from his house to the field is 50 m, tan–1 (3/4) north to east. O Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction. Displacement Distance between final and initial position. a) Vave of plane (Distance/Time) = 260/0.5 = 520 km/hr. b) Vave of bus = 320/8 = 40 km/hr. c) plane goes in straight path 40 m B N 40 m E 50 m S 20 m D 30 m A C E A Initial point (starting point) Y B A X (–20 m, 0) O (20 m, 0) velocity = Vave = 260/0.5 = 520 km/hr. d) Straight path distance between plane to Ranchi is equal to the displacement of bus. Velocity = Vave = 260/8 = 32.5 km/hr. 4. 5. a) Total distance covered 12416 – 12352 = 64 km in 2 hours. Speed = 64/2 = 32 km/h b) As he returns to his house, the displacement is zero. Velocity = (displacement/time) = 0 (zero). Initial velocity u = 0 ( starts from rest) Final velocity v = 18 km/hr = 5 sec (i.e. max velocity) Time interval t = 2 sec. Acceleration = aave = 6. v u 5 t 2 = 2.5 m/s2. In the interval 8 sec the velocity changes from 0 to 20 m/s. change in velocity Average acceleration = 20/8 = 2.5 m/s2 time 2 7. Distance travelled S = ut + 1/2 at 0 + 1/2(2.5)82 = 80 m. In 1st 10 sec S1 = ut + 1/2 at2 0 + (1/2 × 5 × 102) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. From 10 to 20 sec (t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, 3.1 20 Initial velocity u=0 10 8 4 Time in sec 1000 750 S (in ft) 250 0 10 20 30 t (sec) Chapter-3 8. Distance S2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s2. At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S3 = ut + 1/2 at2 = 50 × 10 + (1/2)(–5)(10)2 = 250 m Total distance travelled is 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft. a) Initial velocity u = 2 m/s. final velocity v = 8 m/s t 8 time = 10 sec, acceleration = 6 4 2 v u 82 = 0.6 m/s2 ta 10 b) v2 – u2 = 2aS Distance S = v 2 u2 2a = 8 2 22 2 0 .6 5 t 10 = 50 m. c) Displacement is same as distance travelled. Displacement = 50 m. 9. a) Displacement in 0 to 10 sec is 1000 m. 100 time = 10 sec. Vave = s/t = 100/10 = 10 m/s. 50 b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s. 0 2.5 5 7.5 10 15 at 2 sec. Vinst = 20 m/s. (slope of the graph at t = 2 sec) At 5 sec it is at rest. Vinst = zero. At 8 sec it is moving with uniform velocity 20 m/s Vinst = 20 m/s At 12 sec velocity is negative as it move towards initial position. Vinst = – 20 m/s. 5 m/s 10. Distance in first 40 sec is, OAB + BCD A = 1 1 × 5 × 20 + × 5 × 20 = 100 m. 2 2 Average velocity is 0 as the displacement is zero. B O t D t (sec) 40 20 C 11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m and t = 12 sec s = 20 m So for time interval 0 to 12 sec Change in displacement is zero. So, average velocity = displacement/ time = 0 B 20 10 10 12 20 The time is 12 sec. 12. At position B instantaneous velocity has direction along BC . For average velocity between A and B. Vave = displacement / time = ( AB / t ) t = time 3.2 y 4 B C 2 2 4 6 x Chapter-3 We can see that AB is along BC i.e. they are in same direction. The point is B (5m, 3m). 13. u = 4 m/s, a = 1.2 m/s2, t = 5 sec 1 2 Distance = s = ut at 2 = 4(5) + 1/2 (1.2)52 = 35 m. 14. Initial velocity u = 43.2 km/hr = 12 m/s u = 12 m/s, v = 0 a = –6 m/s2 (deceleration) Distance S = v 2 u2 2( 6) = 12 m 3.3 Chapter-3 15. Initial velocity u = 0 Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks) t = 30 sec v = u + at 0 + 2 × 30 = 60 m/s 1 2 a) S1 = ut at 2 = 900 m when breaks are applied u = 60 m/s v = 0, t = 60 sec (1 min) Declaration a = (v – u)/t = = (0 – 60)/60 = –1 m/s2. S2 = v 2 u2 = 1800 m 2a Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s c) Half the maximum speed = 60/2= 30 m/s Distance S = v 2 u2 2a = 30 2 0 2 22 = 225 m from starting point When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s. u = 60 m/s, v = 30 m/s, a = –1 m/s2 Distance = v 2 u2 30 2 60 2 = = 1350 m 2a 2( 1) Position is 900 + 1350 = 2250 = 2.25 km from starting point. 16. u = 16 m/s (initial), v = 0, s = 0.4 m. Deceleration a = Time = t = v 2 u2 2s v u 0 16 a 320 = –320 m/s2. = 0.05 sec. 17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0 Deceleration = a = v 2 u2 2s = 0 (350)2 = –12.2 × 105 m/s2. 2 0.05 Deceleration is 12.2 × 105 m/s2. 18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec a= v u 50 = 1 m/s2. t 5 1 2 s = ut at 2 = 12.5 m a) Average velocity Vave = (12.5)/5 = 2.5 m/s. b) Distance travelled is 12.5 m. 19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed) Distance travelled in this time is S1 = 15 × 0.2 = 3 m. When brakes are applied, u = 15 m/s, v = 0, a = –6 m/s2 (deceleration) 3.4 Chapter-3 S2 = v 2 u 2 0 15 2 = 18.75 m 2a 2( 6) Total distance s = ...
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