Respuesta :

Answer:

Step-by-step explanation:

Given

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

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

Distance from y axis at

velocity is given by [tex]v=\frac{\mathrm{d} r}{\mathrm{d} t}[/tex]

[tex]v=a\hat{i}+ba\cos (at)\hat{j}[/tex]

Similarly acceleration is given by

[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]

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

Magnitude of acceleration is [tex]=\sqrt{(-a^2b\sin (at))^2}[/tex]

[tex]=a^2b\sin (at)----2[/tex]

From 1 and 2 we can see that

Magnitude of acceleration is proportional to distance from x axis

[tex]a\propto distance\ from\ x-axis[/tex]

Answer:

Step-by-step explanation:

The displacement function is given by

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

Differentiate both sides with respect to t on both the sides

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

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

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

where, A is the acceleration

So, by equation (1)

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

So, the acceleration is proportional to the displacement function.