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Question #43481

Difficulty: 8/10 | 🟢 Free

The Question

A particle of mass mm is initially at rest. From t=0t = 0, it is driven in a straight line by an engine that delivers a constant power PP. The particle experiences a resistive drag force Fd=kvF_d = kv, where vv is the instantaneous velocity and kk is a positive constant. Which of the following expressions correctly represents the velocity vv of the particle as a function of time tt?

Options

  • A.

    Pk(1−e−2kt/m)\sqrt{\frac{P}{k} \left(1 - e^{-2kt/m}\right)}

  • B.

    Pk(1−e−kt/m)\sqrt{\frac{P}{k} \left(1 - e^{-kt/m}\right)}

  • C.

    Pk(1−e−kt/m)\sqrt{\frac{P}{k}} \left(1 - e^{-kt/m}\right)

  • D.

    Pk(1−e−2kt/m)\sqrt{\frac{P}{k}} \left(1 - e^{-2kt/m}\right)

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Non-Uniform Acceleration Constraint|Power-Velocity Relationship|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to transition from constant-acceleration kinematics to calculus-based dynamics under a power constraint.

A common trap is incorrectly assuming that constant power implies a constant driving force, or attempting to apply standard kinematic equations.

Furthermore, students who correctly set up the differential equation m(dv/dt)=(P/v)−kvm(dv/dt) = (P/v) - kv often mismanage the integration of ∫v/(P−kv2)dv\int v/(P-kv^2)dv.

Missing the factor of 1/21/2 from the chain rule during u-substitution leads directly to the incorrect exponential decay rate in Option B, effectively filtering out students with weak algebraic manipulation skills in physics contexts.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Straight Line - Class 11 Part 1
ConceptNon-Uniform Acceleration Constraint
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree