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#velocity calculation

9 public questions tagged with this topic.

Motion Parameters and Calculations

A particle’s position is given by x\=2t2+3t and y\=t2−2t (in meters and seconds). What is the magnitude of its velocity

Velocity: vx=dxdt=4t+3,vy=dydt=2t−2. 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/s 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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Straight Line Motion - Mixed Concepts

A ball is thrown at 30m/s at 53∘. What is its vertical velocity at t\=2s? (Take g\=10m/s2,sin⁡53∘\=0.8)

Vertical velocity vy=v0sin⁡θ0−gt. 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/s as the result, so option C is correct.

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

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Motion in a Straight Line - Advanced Problems

A stone is dropped from rest. What is its velocity after falling 24.5m? (Take g\=10m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 25 m/s. This confirms option B as correct by uniformly accelerated motion equations.

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 stone is thrown upwards with a speed of 14m/s. What is its velocity after 1.5s? (Take g\=10m/s2)

Use v=v0+at. Here, v0=14m/s, a=−10m/s2, t=1.5s. 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 -1 m/s as the result, so option B 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 stone is dropped from rest. What is its velocity after falling 78.4m? (Take g\=9.8m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 40 m/s. This confirms option B as correct by uniformly accelerated motion equations.

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 ball is dropped from a height of 80m. What is its velocity just before hitting the ground? (Take g\=10m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 40 m/s. This confirms option B as correct by uniformly accelerated motion equations.

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

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