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

  • Thread starter cj
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  • #1
cj
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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:

Answers and Replies

  • #2
ehild
Homework Helper
15,498
1,878

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.
 
  • #3
cj
82
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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.
Hmm... trying to follow your advice.
click here
The stated answer is 6 m/s2, which I'm not finding any way to get. Maybe 6 m/s2 is incorrect?
 
  • #4
haruspex
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Insights Author
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Hmm... trying to follow your advice.
click here
The stated answer is 6 m/s2, which I'm not finding any way to get. Maybe 6 m/s2 is incorrect?
The right hand side of the first line of step 6 is wrong. How do you differentiate ##\sqrt{x^2+y^2}##?
An easier way is to start with ##|v|^2=\vec v.\vec v## and differentiate both sides.
 
  • #5
cj
82
0
The right hand side of the first line of step 6 is wrong. How do you differentiate ##\sqrt{x^2+y^2}##?
An easier way is to start with ##|v|^2=\vec v.\vec v## and differentiate both sides.
Ah! Got it!! Thank you so much.
 

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