Respuesta :
Answer:
Velocity at the point of maximum x cordinate is 9.578m/s
Position vector of the particle when it reaches point of maximum x ordinate is [tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]
Explanation:
We shall resolve the motion of the particle along x and y direction separately
The particle will reach it's maximum x coordinate when it's velocity along x axis shall become 0
We have acceleration along x-axis = [tex]-2.6m/s^{2}[/tex]
acceleration along y-axis = [tex]4.7m/s^{2}[/tex]
Thus using the first equation of motion along x axis we get
[tex]v_{x}=u_{x}+a_{x}t\\\\[/tex]
Applying values we get
[tex]0=5.3-2.6t\\\\\therefore t=\frac{5.3}{2.6}sec\\\\t=2.038sec[/tex]
Now to obtain it's position we shall use third equation of motion
[tex]v_{x}^{2}=u_{x}^{2}+2as_{x}\\\\0=(5.3)^{2}+2(-2.6)s_{x}\\\\\therefore s_{x}=\frac{-28.09}{-5.2}m\\\\s_{x}=5.402m[/tex]
Now it's location along y- axis can be obtained using 2nd equation of motion along the y axis
[tex]s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}[/tex]
Applying values as follows we get
[tex]u_{y}=0\\a_{y}=4.7m/s^{2}\\t=2.038s[/tex]
[tex]s_{y}=0\times 2.038+\frac{1}{2}\times 4.7m/s^{2}\times2.038^{2}\\\\s_{y}=9.76m[/tex]
thus the position vector of the particle when it reaches it's maximum x co-ordinate is
[tex]\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}[/tex]
Now velocity of the particle at the position of maximum x co-ordinate shall be zero along x-axis and along the y-axis it can be found along the first equation of motion along y axis
[tex]v_{y}=u_{y}+a_{y}t\\\\v_{y}=0+4.7\times 2.038\\\\v_{y}=9.578m/s[/tex]
