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A particle moves so that r(t) = ati + b sin atj. Show that the magnitude of the acceleration of the particle is proportional to its distance from the x-axis.

User Tkyass
by
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2 Answers

3 votes

Answer:

Explanation:

The displacement function is given by


\overrightarrow{r(t)}=at\widehat{i}+bSin(at)\widehat{j} .... (1)

Differentiate both sides with respect to t on both the sides


\overrightarrow{v}=\frac{\overrightarrow{r(t)}}{dt}=a\widehat{i}+abCos(at)\widehat{j}

Differentiate again with respect to t to get the function of acceleration


\overrightarrow{A}=\frac{\overrightarrow{v(t)}}{dt}=-a^(2)bSin(at)\widehat{j}

where, A is the acceleration

So, by equation (1)


\overrightarrow{A}=\frac{\overrightarrow{r(t)}-at\widehat{i}}{a^(2)}

So, the acceleration is proportional to the displacement function.

User Badunk
by
8.6k points
4 votes

Answer:

Explanation:

Given

Position of particle is
r(t)=at\hat{i}+b\sin (at)\hat{j}

i.e. distance from x axis is
b\sin (at)---1

Distance from y axis at

velocity is given by
v=\frac{\mathrm{d} r}{\mathrm{d} t}


v=a\hat{i}+ba\cos (at)\hat{j}

Similarly acceleration is given by


a=\frac{\mathrm{d} v}{\mathrm{d} t}


a=0\hat{i}-a^2b\sin (at)\hat{j}

Magnitude of acceleration is
=√((-a^2b\sin (at))^2)


=a^2b\sin (at)----2

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis


a\propto distance\ from\ x-axis

User EricL
by
8.1k points

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