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## Homework Statement

At a moment in time a particle has a velocity of v = 3 m/s

(better formatted reference: click)

**î**+ 4 m/s**ĵ**, and an acceleration of a = 6 m/s^{2}**î**+ 3 m/s^{2}**ĵ**. Find the rate of change of the speed of the particle, that is, find d|**v**|/dt.(better formatted reference: click)

The answer is: 6 m/s

^{2}.**How on Earth????**## Homework Equations

speed = magnitude of velocity

a = dv/dt

a = dv/dt

## The Attempt at a Solution

The magnitude of the initial velocity (the speed) is 5 m/s. 3^2 + 4^2 = 5^2.

Acceleration is constant, so I tried using the equations of uniform acceleration kinematics, e.g. v

Can't, for the life of me, see how the rate of change of speed = 6m/s

Thanks for any hints.

Acceleration is constant, so I tried using the equations of uniform acceleration kinematics, e.g. v

_{x}=∫(a_{x})dt, etc.Can't, for the life of me, see how the rate of change of speed = 6m/s

^{2}.Thanks for any hints.

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