Distance travelled by the particle

[utkarshakash]

Homework Statement

A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from v1 to v2 is(neglect friction)

Homework Equations

Work done = ΔKE
P=dW/dt
s=(v2^2-v1^2)/2a

The Attempt at a Solution

Since work done by all forces equals ΔKE
∴ Work done = m(v2^2-v1^2)/2
Now P=dW/dt,
differentiating W wrt to t, I get
a=P/m(v2-v1)
Putting a in last equation I get
s=m(v2^2-v1^2)(v2-v1)/2P

but this answer is wrong :(

The correct answer given in my book is m(v2^3-v1^3)/3P
I don't know where I've commited the mistake.

 

[PeterO]

Homework Statement

A constant power P is applied to a particle of mass m. The distance traveled by the particle when its velocity increases from v1 to v2 is(neglect friction)

Homework Equations

Work done = ΔKE
P=dW/dt
s=(v2^2-v1^2)/2a

The Attempt at a Solution

Since work done by all forces equals ΔKE
∴ Work done = m(v2^2-v1^2)/2
Now P=dW/dt,
differentiating W wrt to t, I get
a=P/m(v2-v1)
Putting a in last equation I get
s=m(v2^2-v1^2)(v2-v1)/2P

but this answer is wrong :(

The correct answer given in my book is m(v2^3-v1^3)/3P
I don't know where I've commited the mistake.

Being a cynic about answers like this given in texts, I would consider an equivalent problem with numerical values and check that the "formula" answer does give the correct answer. Perhaps your answer does also, and there is just 2 ways of saying the same thing.

Use values that should not give a coincidental answer ie avoid a value of 2, since 2 + 2 and 2 x 2 give the same answer.

EG 14 kg mass traveling at 7 m/s, is accelerated to 11 m/s when power is added at the rate of 5 watts. How far would be covered.

Your formula says 14(121 - 49)/10
Their Formla says 14(1331 - 343)/15

They are not the same answer, so at least one of them is wrong.

Hint: find the initial and final Kinetic energies. Since the energy was added at a rate of 5 J/s, you can work out how long in time that takes.
Since the average speed is 9m/s, you can thus calculate the distance covered.
While doing that numerical calculation, you may even see why your/their answer will be correct.

Good luck.
Note: You could repeat the calculation for some different values, to show your agreement was not coincidental.