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Physics NEET MCQ
Motion In One Dimension Mcq
Quiz 1
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Q.1
A body starts from rest and travels 120cm in the 8th second, then acceleration of the body is .. [AFMC1997]
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a) 1.02m/s2
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b) 0.34 m/s2
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c) 0.18 m/s2
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d) 0.16 m/s2
Explanation
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)
Q.2
A car travels first half the distance between two places with a speed of 30km/hr and the remaining half with a speed of 50km/hr. The average speed of the car is ..
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a) 45 km/hr
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b) 42.8 Km/hr
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c) 37.5 Km/hr
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d) 48 Km/hr
Explanation
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)
Q.3
The displacement x of a particle moving along a straight line at time t is given by x=a0+a1t+a2t2 The acceleration of the particle is
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a) 4a2
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b) 2a2
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c) 2a1
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d)a2
Explanation
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)
Q.4
Find the total displacement of a body in 8 seconds starting from rest with an acceleration of 20cm/sec2[AFMC 2000]
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a) 64m
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b) 64cm
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c) 640cm
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d) 0.06m
Explanation
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)
Q.5
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 +12t - t3 how far would the particle be before coming to rest?[AFMC 2000]
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a) 24 m
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b) 40 m
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c) 56 m
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d) 16 m
Explanation
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)
Q.6
A particle covers half of its total distance with speed v1 and the rest half distance with speed vIts average speed during the complete journey is .. [ CBSE PMT 2011]
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a)
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b)
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c)
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d)
Explanation
Let total distance traveled by the particle be 2s ThenAnswer: (b)
Q.7
Two spheres of same size one of mass 2 kg and another of mass4 kg are dropped simultaneously from the top of Qutab Minar ( height 72m). When they are 1m above the ground, the two spheres have same.. [ AIIM 2006]
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a) momentum
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b) kinetic energy
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c) potential energy
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d) acceleration
Explanation
gravitational acceleration is independent of mass Answer: (d)
Q.8
A bus start from rest with an acceleration of 1 m/secA man who is 48 meter behind the bus with a uniform velocity of 10 m/sec. then the minimum time after which the man will catch the bus is [AFMC 2001]
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a) 4 sec
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b) 10 sec
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c) 12 sec
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d) 8 sec
Explanation
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)
Q.9
The displacement of a particle moving in straight line depends on time as x=αt3 + βt2 + γt + δThe ratio of initial acceleration to its initial velocity depends [AFMC 2002]
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a) only on α and γ
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b) only on β and γ
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c) only on α and β
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d)only on α
Explanation
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)
Q.10
A particle covers 150 m in 8th second starting from rest, its acceleration is [AFMC2003}
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a) 15m/s2
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b) 20 m/s2
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c) 10 m/s2
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d) 8 m/s2
Explanation
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)
Q.11
A coin is dropped in lift. It takes a time t1 to reach the floor when lift is stationary. It takes time t2 when lift is moving up with constant acceleration. Then [AFMC2005}
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a) t1 >t2
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b) t2>t1
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c) t1=t2
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d) t1>>t2
Explanation
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)
Q.12
A person is standing between two cliffs. A sound is produced, the person hears echo after 3 seconds and 5 seconds. Velocity of sound in air=336m/s. the separation between two cliffs is [AFMC 2006]
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a) 1344m
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b) 2688m
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c) 772m
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d) none of these
Explanation
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)
Q.13
A car is moving with uniform acceleration. it covers 200 m in 2 sec and 220 m in next 4 sec. The velocity of car at the end of 7th second from start is .
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a) 1m/s
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b) 0.1m/s
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c) 10m/s
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d) 20m/s
Explanation
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)
Q.14
the area of the acceleration displacement curve of a body gives [AFMC2006]
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a) impulse
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b) change in momentum per unit mass
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c) change in KE per unit mass
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d) total change in energy
Explanation
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)
Q.15
A train of 150 meter long is going towards north direction at speed of 10m/s. A parrot flies at the speed of 5m/s to wards south direction parallel to the railway track. The time taken by the parrot to cross the train is.. [ CBSE-PMT 1992]
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a)12 sec
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b) 8 sec
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c)15 sec
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d)10 sec
Explanation
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)
Q.16
A bus traveling the first one third distance at a speed of 10 km/h, the next one third at 20 km/h and the last one-third at 60 km/h. The average speed of the bus is ... [ CBSE-PMT 1991]
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a) 9 km/h
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b) 16 km/h
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c)18 km/h
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d)48 km/h
Explanation
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)
Q.17
A car accelerates from rest at a constant rate α for some time, after which it decelerated at constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is .. [ CBSE-PMT 1994]
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a)
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b)
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c)
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d)
Explanation
From graph Vmax=AD=αt1=β t2 But t=t1 + t2 t=(Vmax / t1) + (Vmax / t2) Answer:(d)
Q.18
A particle moves along a straight line such that its displacement at any time 't' is given bys=( t3-6t2+3t+4) metersThe velocity when the acceleration is zero is ... [ CBSE PMT 1994]
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a) 3 m/s
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b) -12 m/s
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c) 42 m/s
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d) -9 m/s
Explanation
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)
Q.19
Three different objects of mass m1 , m2, m3 are allowed to fall from rest and from the same point O along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of .. [ CBSE-PMT 1995]
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a)m1 : m2: m3
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b) m1 : 2m2: 3m3
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c)1:1:1
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d)1/m1 : 1/m2: 1/m3
Explanation
Path is frictionless and objects are falling freely due to gravity from same height, thus all the masses have same velocityAnswer: (c)
Q.20
A particle has initial velocity and acceleration respectively. Its speed after 10 s is ... [ CBSE - PMT 2010]
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a) 7 unit
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b) 7√2 unit
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c)8.5 unit
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d)10 unit
Explanation
using equation of motion in vector form we get|v|=7√2 unit Answer:(b)
Q.21
A ball is dropped form a high rise platform at t=0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t=18s. What is the value of v? (g=10m/s2) [ CBSE-PMT 2010]
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a) 75 m/s
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b) 55m/s
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c) 40 m/s
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d) 60 m/s
Explanation
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)
Q.22
A bus is moving with a speed of 10 m/s on a straight road. A scooterist wishes to overtake the bus in 100s. If the bus is at a distance of 1km from the scooterist, with what speed should the scooterist chase the bus? [ CBSE PMT 2009]
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a)40 m/s
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b) 25 m/s
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c)10 m/s
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d)20 m/s
Explanation
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)
Q.23
A particle moving in a straight line with a constant acceleration. It changes its velocity from 10 m/s to 20m/s, while passing through a distance 135 m in t second. The value of t is ... [ CBSE-PMT 2008]
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a) 10
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b) 1.8
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c)12
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d)9
Explanation
average velocity=(10+20) / 2=15m/sdisplacement=average velocity × time135=15 × t t=135/15=9 secAnswer: (d)
Q.24
A man of 50kg mass is standing in a gravity free space at height of 10 m above the floor. he throws a stone of 0.5kg mass downwards with speed of 2 m/s. When the stone reaches the floor the distance of the man above the floor will be ... [ CBSE-PMT 2010]
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a) 9.9 m
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b) 10.1 m
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c)10.0
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d)20 m
Explanation
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)
Q.25
A particle starts its motion from rest under the action of a constant force. if the distance covered in first 10 second is S1 and that covered in the first 20 seconds is S2, then .. [ CBSE-PMT 2009]
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a) S2=3S1
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b) S2=4S1
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c) S2=S1
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d) S2=2S1
Explanation
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)
Q.26
A body starts from rest, what is the ratio of the distance traveled by the body during the 4th and 3rd second? [ CBSE-PMT 1993]
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a)7/5
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b) 5/7
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c)7/3
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d)3/7
Explanation
Displacement in nth second is given by formula Answer: (a)
Q.27
A boy standing at the top of a tower of 20m height drops a stone. Assume g=10 m / s2, the velocity with which it hits the ground is .. [ CBSE-PMT 2011]
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a) 10.0 m/s
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b) 20.0 m/s
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c)40.0 m/s
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d)5.0 m/s
Explanation
We have V2=u2 + 2gh here u=0 thus V=√(2gh) V=√(×10×20)=20 m/sAnswer: (b)
Q.28
A body is moving with velocity 30 m/s towards east. After 10 sec. its velocity becomes 40 m/s towards north . The average acceleration of the body is .. [ CBSE-PMT 2011]
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a)1 m/s2
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b)7 m/s2
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c)10 m/s2
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d)5 m/s2
Explanation
As shown in figure initial velocity u=30 i and final velocity v=40 j
=5 m/ s2 Answer:(d)
Q.29
A particle moves a distance x in time 't' according to equation x=( t + 5 ) -1 . the acceleration of particle is proportional to .. [ CBSE-PMT 2010]
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a) (velocity) 3/2
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b) (distance) 2
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c) (distance) -2
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d) (velocity) 2/3
Explanation
velocity v=dx/dt=(-1) (t+5)-2 Thus (t+5)=-v-1/2 acceleration, a=dv/dt=(2) (t+5)-3 on substituting value of (t+5) from above in equation of acceleration we get a=-2v3/2Answer: (a)
Q.30
A car moving with a speed of 40 km/h can be stopped by applying brakes at least after 2m. If the same car is moving with a speed of 80km/h what is the minimum stopping distance? [ CBSE - PMT1998]
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a)8 m
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b) 6 m
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c)4 m
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d)2 m
Explanation
According to formula v2=u2 + 2asin both the cases v=0thus a=- u2 / 2s Retardation is constant thus s ∝ u2 S2=8 mAnswer: (a)
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