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#position functions

3 public questions tagged with this topic.

Straight Line Motion - Mixed Concepts

A particle’s position is given by x\=5t−t2 and y\=2t2 (in meters and seconds). What is the magnitude of its acceleration

Velocity: vx=dxdt=5−2t,vy=dydt=4t. 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 2 m/s² as the result, so option A is correct.

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

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Equations of Motion and Problem Solving

A particle’s position is given by x\=2t+t2 and y\=3t−2t2 (in meters and seconds). What is its speed at t\=2s?

Velocity: vx=dxdt=2+2t,vy=dydt=3−4t. 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 6 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

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Equations of Motion and Problem Solving

A particle’s position is given by x\=3t2−4t and y\=5t (in meters and seconds). What is its speed at t\=1s?

Velocity: vx=dxdt=6t−4,vy=dydt=5. 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 A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving

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