Question

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.

Answer 1

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.

Answer 2

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`