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Salsk061 [2.6K]
3 years ago
12

Solve for m2= F ---- m

Physics
1 answer:
Pie3 years ago
3 0
Don't think m has a
value
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B) It does not deflect at all

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If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w
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\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath

The velocity at time t is

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath

Take two vectors that point in the positive x and positive y directions, such as \vec\imath and \vec\jmath. The dot products of the velocity vector with \vec\imath and \vec\jmath are

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and

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta

We want the angles between these vectors to be 45º, for which we have \cos45^\circ=\frac1{\sqrt2}. So

\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0

\implies t(3ct-2b)=0

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When t=0, the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

\boxed{t=\dfrac{2b}{3c}}

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