According to formula for displacement in nth second Here Sn=120cm=1.2m time n=8 sec initial velocity u=0 ( given) Substitute above values in equation and solve for 'a' we get a=0.16m /s2 Answer=(d)

Let the distance be S for first half distance Let time taken to cover the distance S be t hence t=distance / average speed t=S/30 hr Similarly time taken to travel second half t'=S/50 hr Average speed=(Total distance) /(total time)--eq(1) Total distance=S+S=2S Total time=t+t'=(S/30) + (S/50) Substituting the values of total distance and total time in equation(1) we get Average speed=37.5 km/h Answer: (c)

Here displacement is the function of time. By taking second derivative of x with respect to time 't' we will get the formula for acceleration 'a' Answer: (b)

formula for displacement Here initial velocity u=0 time t=8sec acceleration a=20cm/sec2 By substituting above values in equation for displacement, and solving equation for S we get S=640cm Answer: (c)

Take the first order derivative to get formula of velocity equate formula of velocity with zero. To get the value of t 0=12-3t2 t=2sec Substitute value of t in equation for distance x=40 + 12(2) - (2)3 X=56 m, Particle will be at 56m from O Note that question is "how far would the particle be before coming to rest?" and not about dispacement in time interval 0 to 2sec Thus distance should from "O" Answer:(c)

Let total distance traveled by the particle be 2s ThenAnswer: (b)

gravitational acceleration is independent of mass Answer: (d)

Let the distance traveled by the bus when man catch the bus be S meter distance traveled by the man when he catch the bus S'=S+48 meter Initial velocity of bus u=0 (given) Initial velocity of man u'=10m/sec (given) Acceleration of Bus=1 m/s2 (given) Acceleration of Man=0 (given) We will use formula for displacement S=ut+½(t2)By substituting the values in above equation we get Displacement of Bus S=½(t2) -eq(1) Displacement of Man S+48=10t -eq(2) By substituting the value of S from eq(1) in eq(2) we get ½(t2)+48=10 t After solving above Quadratic equation we get t=12 sec and t=8 sec Minimum Value of t=8 sec Answer:(d)

Displacement is function of time, If we take first order derivative of above equation we will get formula for velocity By substituting t=0 in above equation we will get value of initial velocity V=γ If we take second order derivative of equation for displacement we will get formula for acceleration By substituting t=0 in above equation we will get value of initial acceleration a=2β Ratio of initial acceleration to initial velocity=2β / γ Answer:(b)

According to formula for displacement in nth second By substituting the values of initial speed u=0 time n=8 we get On simplification we get a=20 m/s2 Answer:(b)

Lift is accelerated frame of reference as it is given that lift is moving up with constant ACCELERATION. Therefore coin will experience pseudo downward force and resultant acceleration will be greater than the gravitational acceleration Therefore time taken by coin to reach the floor will be less than the stationary lift Answer:(a)

When we hear echo distance traveled by the sound wave is two times the actual distance between the wall or cliff. Hence time taken by the sound to reach the wall is half of the time of echo heard.Now total time for first and second echo is=3sec +5 secThus time taken by the sound to reach cliff=8sec/2=4 secDistance between the cliff=time taken by sound to reach the cliff × velocity of sound Distance between the cliff=(4) × (336)Distance between the cliff=1344 m Answer: (a)

It is given acceleration is uniform let it be 'a' To find the velocity we will find first what is the value of 'a' Let initial velocity be u According to formula for displacement S=ut+½(t2) i) Distance covered in first 2 sec=200m Substituting above values in the equation for displacement we get 200=u(2) + ½a(22) on simplification we get 100=u+a -eq(1) ii) Distance covered in next 4 sec=220 m So distance traveled in 6 sec=200+220=420 m Substituting value of time t=6sec and displacement=420m in the equation for displacement we get 420=u(6) +½a(62) 70=u+3a -(eq(2) Solving equation 1 and 2 we get u=115 m/s and a=-15 m/s2) Substituting avove values in equation V=u + at v=115 -15(7) v=115 - 105 V=10 m/s Answer: (c)

Area under the curved can be calculated by the mathematical tool "Integration" Let displacement is along X-axis and acceleration along Y axis For small displacement of dx consider acceleration as constant, then we get formula for small area dA dA=adx Now let the particle is displaced from position X1 to X2 Then A=∫adx Let velocity of particle at X1 be u and at X1 be v Also acceleration 'a' is first order derivative of velocity Substituting the value of acceleration in terms of velocity and changing the upper limit and lower limit to v and u respectively we get multiplying dividing above equation by mass 'm' we get A=Change in kinetic energy per unit mass Answer:(c)

Let v be the velocity of parrot w.r.t observer=5 m/s let v' be the velocity of train w.r.t. observer=-10 m/sNow velocity of parrot with respect to train=v - v' Relative velocity of parrot w.r.t train=5-(-10)=15 m/s Time taken by parrot to cross the train=150 /15=10sAnswer: (d)

Let the distance be 3sTime taken to travel first one third distance s=s/10 hoursTime taken to travel second one third distance s=s/20 hours Time taken to travel third one third distance s=5/60 hoursTotal time taken=s/10 + s/20 + s/60=s/6 hoursAverage speed=Total distance / total time Average speed=3s /(s/6)=18 km/hAnswer: (c)

From graph Vmax=AD=αt1=β t2 But t=t1 + t2 t=(Vmax / t1) + (Vmax / t2) Answer:(d)

Velocity v=ds /dt=3t2 -12t + 3 Acceleration a=dv/ dt=6t -12 For a=0 we have 0=6t -12 or t=2s. Hence at t=2 sec velocity v=3×(2)2 -12(2)+3 V=-9 m/s Answer: (d)

Path is frictionless and objects are falling freely due to gravity from same height, thus all the masses have same velocityAnswer: (c)

using equation of motion in vector form we get|v|=7√2 unit Answer:(b)

First ball initial speed u=0 distance traveled in 18ses S=½at2 S=½(10)(18)2 --eq(1) Second Ball : Second ball must traveled same distance of as of first to meet in 12 sec, thus S=u'(t) +½at2 u'=v , time for second ball=18-6=12 sec S=v(12) + ½(10)(12)2 --eq(2) From equation 1 and 2 we get ½(10)(18)2=v(12) + ½(10)(12)2 on solving equation for 'v' v=75 m/s Answer: (a)

Distance traveled by bus=velocity × time=10 × 100=1000 mInitial distance between scooterist and bus=1000 mtotal distance to be traveled by scooterist=2000 mtime for scooterist to catch bus=100 secvelocity of scooterist=2000/100=20 m/sAnswer: (d)

average velocity=(10+20) / 2=15m/sdisplacement=average velocity × time135=15 × t t=135/15=9 secAnswer: (d)

No external force acts there fore according to law of conservation of momentumMomentum before throwing the stone=momentum of stone + momentum of man0=50u + 0.5×2where u is the velocity of man u=(-1/50) m/s moving away from floorTime taken by the stone to reach flooring=10/2=5 secDisplacement of man in upward direction in 2 sec=(1/50)×5=0.1 m∴ when stone reaches the floor, the distance between the man and floor=10.1m Answer:(b)

given u=0, t1=10s, t2=20s, acceleration is same in both the cases Using the relation S=ut + ½ at2S1=½ a × t12 S2=½ a × t22 ∴ S1 / S2=( t1 / t2) 2S1 / S2=( 10 /20)2S2=4S1 Answer: (b)

Displacement in nth second is given by formula Answer: (a)

We have V2=u2 + 2gh here u=0 thus V=√(2gh) V=√(×10×20)=20 m/sAnswer: (b)

As shown in figure initial velocity u=30 i and final velocity v=40 j =5 m/ s2 Answer:(d)