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.
