# Block at half its final speed

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A block of mass M starting from rest slides down a frictionless inclined plane of length L. When the block has attained 1/2 its final speed, the distance it has traveled along the plane is

choices: L/4, L/2, L/sqrt(2), 3L/4

I tried using the equation V2=v02+2ax and set v0 = 1/2v which becomes 1/4 v2 to get 3/4 v2 = 2ax and then 3a/8 v2 = x

But I don't think I'm doing it right.

What should I be doing?

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What's the relation between v and x in your equation? What shape would a graph be?

Hint: by saying look for the shape of a graph and the relation I mean that you concentrate on the fundamentals - the initial velocity is zero the acceleration is constant so you can basically ignore them

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Three hints in the form of questions

At what distance travelled does the block reach its final speed?

In what direction do we take the velocity be for the purpose of the equations of motion?

Velocity is a vector do we need to resolve it in any particular directions?

Edited by studiot
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A block of mass M starting from rest slides down a frictionless inclined plane of length L. When the block has attained 1/2 its final speed, the distance it has traveled along the plane is

choices: L/4, L/2, L/sqrt(2), 3L/4

I tried using the equation V2=v02+2ax and set v0 = 1/2v which becomes 1/4 v2 to get 3/4 v2 = 2ax and then 3a/8 v2 = x

But I don't think I'm doing it right.

No, you aren't. v0 isn't 1/2v, since it starts at rest. v0 = 0

But you could use that equation to determine the speed at L, and then solve for half that speed. Using v0 = 0, the equation becomes vf2 = 2ax

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Thanks

v2 = 2ax

x = L

So then 1/2 of that final v = 1/2 sqrt(2aL)

And the equation for distance is now just x = 1/2 at2 So we need to find a or t

So a = (v-v0)/t and I then simplified that into t2=1/2 L/a

And plugged that back into the distance equation to get 1/4 L.

I think that's right?

Edited by rasen58
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Thanks

v2 = 2ax

x = L

So then 1/2 of that final v = 1/2 sqrt(2aL)

And the equation for distance is now just x = 1/2 at2 So we need to find a or t

So a = (v-v0)/t and I then simplified that into t2=1/2 L/a

And plugged that back into the distance equation to get 1/4 L.

I think that's right?

Or use the same equation. 1/2 vf = sqrt(2aL)/2, as you say. So put that into the equation v2 = 2ax (v0 is still 0)and solve for x

2aL/4 = 2ax

x = L/4

Intuitively you might expect this, in terms of potential and kinetic energy. An object that doubles its speed has 4x the energy, so it has to travel 4x as far.

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• 6 years later...

yes use that equation!

1). set v to any letter, i chose v. set v= 0 because it starts from rest. plug in L for Δx. solve for a. i got a = v2 / 2L.

2). using the V2=v02+2ax equation, set v = 1/2v, v0 = 0, and a = v2 / 2L. solve for Δx. i got Δx = L/4.

hope this helps!

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3 minutes ago, andrea364499 said:

hope this helps!

He is going to get points off since his homework is 6 years late.

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10 hours ago, Bufofrog said:

He is going to get points off since his homework is 6 years late.

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12 minutes ago, swansont said:

Well that's a relief.🙂

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