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

Answer:

Step-by-step 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 $=\sqrt{(-a^2b\sin (at))^2}$

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

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

$a\propto distance\ from\ x-axis$

Answer 2

Answer:

Step-by-step 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.