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#displacement

12 public questions tagged with this topic.

Young's Modulus and Elasticity

A brass block of dimensions 0.5m×0.3m×0.2m is subjected to a shearing force of 6×104N. If the shear modulus of brass is

Shear modulus: G = F/AΔx/L. Rearrange: Δx = FLAG. Area: A = 0.5×0.3 = 0.15m2, L = 0.2m. Substitute: Δx = 6×104×0.20.15×3.6×1010 = 120005.4×109≈2.22×10−6m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.22×10−6m. 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 Parameters and Calculations

A particle has an initial velocity of 3 î m/s and a constant acceleration of 2 ĵ m/s². What is the magnitude of its disp

Displacement r = v₀ t + (1/2) a t². 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 9 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations

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Acceleration and Velocity Analysis

A particle starts with velocity 5i^m/s and accelerates at (−3j^)m/s2. What is its displacement magnitude after 2s?

Displacement r=v0t+12at2. 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 10 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

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Acceleration and Velocity Analysis

A particle starts from rest with an acceleration of (3i^+5j^)m/s2. What is its displacement magnitude after 2s?

Displacement r=12at2 (since v0=0). 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 10 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

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Motion in a Straight Line - Advanced Problems

A stone is thrown upwards at 35m/s from a 45m cliff. How far below the cliff’s edge is it after 8s? (Take g\=10m/s2)

Displacement: y=35⋅8−12⋅10⋅(8)2=280−320=−40m. 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 85 m as the result, so option A 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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Free Fall and Vertical Motion

A car moves at 16m/s and decelerates at 2.5m/s2 for 3s, then accelerates at 3.5m/s2 for 4s. What is the net displacement

Phase 1: v=16−2.5⋅3=8.5m/s, x1=16⋅3−12⋅2.5⋅(3)2=48−11.25=36.75m. 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 95 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion

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Free Fall and Vertical Motion

A stone is thrown upwards at 30m/s from a 55m tower. How far below the tower’s top is it after 7s? (Take g\=10m/s2)

Displacement: y=30⋅7−12⋅10⋅(7)2=210−245=−35m. 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 35 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion

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Kinematic Equations and Uniformly Accelerated Motion

A car moves at 12m/s and decelerates at 2m/s2 for 3s, then accelerates at 4m/s2 for 4s. What is the net displacement?

Phase 1: v=12−2⋅3=6m/s, x1=12⋅3−12⋅2⋅(3)2=36−9=27m. 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 83 m as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion

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Kinematic Equations and Uniformly Accelerated Motion

A car moves at 18m/s and decelerates at 2m/s2 for 4s, then accelerates at 3m/s2 for 5s. What is the net displacement?

Phase 1: v=18−2⋅4=10m/s, x1=18⋅4−12⋅2⋅(4)2=72−16=56m. 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 140 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Kinematic Equations and Uniformly Accelerated Motion

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