# "speed rate of change" vs. "velocity rate of change"

## Homework Statement

At a moment in time a particle has a velocity of v = 3 m/s î + 4 m/s ĵ, and an acceleration of a = 6 m/s2 î + 3 m/s2 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/s2. How on Earth????

## Homework Equations

speed = magnitude of velocity
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. vx=∫(ax)dt, etc.
Can't, for the life of me, see how the rate of change of speed = 6m/s2.
Thanks for any hints.​

Last edited:

ehild
Homework Helper

## Homework Statement

At a moment in time a particle has a velocity of v = 3 m/s î + 4 m/s ĵ, and an acceleration of a = 6 m/s2 î + 3 m/s2 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/s2. How on Earth????

## Homework Equations

speed = magnitude of velocity
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. vx=∫(ax)dt, etc.
Can't, for the life of me, see how the rate of change of speed = 6m/s2.
Thanks for any hints.​

Write the velocity as function of time, determine the speed as function of time, then differentiate. That is the rate of change of speed
at time t. You need the rate of change at t=0.

Write the velocity as function of time, determine the speed as function of time, then differentiate. That is the rate of change of speed
at time t. You need the rate of change at t=0.
The stated answer is 6 m/s2, which I'm not finding any way to get. Maybe 6 m/s2 is incorrect?

haruspex
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