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#distance

26 public questions tagged with this topic.

Vector Product and Rotational Kinematics

Two particles of masses 3kg and 7kg are at (1,4) and (9,2) respectively. What is the distance of their center of mass fr

X = (3×1)+(7×9)3+7 = 3+6310 = 6.6. Y = (3×4)+(7×2)3+7 = 12+1410 = 2.6. Distance = (6.6)2+(2.6)2 = 43.56+6.76 = 50.32≈7.09m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.09 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Two particles of masses 8kg and 2kg are at (0,5) and (10,0) respectively. What is the distance of their center of mass f

X = (8×0)+(2×10)8+2 = 2010 = 2. Y = (8×5)+(2×0)8+2 = 4010 = 4. Distance = (2)2+(4)2 = 4+16 = 20≈4.47m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.47 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Motion in a Straight Line - Advanced Problems

A body accelerates uniformly from rest to a velocity of 20m/s over a distance of 50m. What is the acceleration?

Use v2=v02+2ax. Here, v0=0, v=20m/s, x=50m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s² as the result, so option C is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems

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Distance, Displacement and Velocity-Time Relations

A particle moves at 18m/s and decelerates at 4m/s2 for 2s, then accelerates at 3m/s2 until its speed is 18m/s. What is t

Phase 1: v=18−4⋅2=10m/s, x1=18⋅2−12⋅4⋅(2)2=36−8=28m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 65 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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Distance, Displacement and Velocity-Time Relations

A person walks 10m in 2s and returns to the starting point in 2s. What is the average speed?

Average speed = total distance / total time. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 0 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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Distance, Displacement and Velocity-Time Relations

A train moves with a constant speed of 108km/h for 2min. What is the distance traveled?

Convert speed: 108km/h=108⋅10003600=30m/s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 30 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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Distance, Displacement and Velocity-Time Relations

A body moving at 15m/s decelerates uniformly to rest in 5s. What is the distance covered?

Find a=v−v0t=0−155=−3m/s2. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 25 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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