Question

A particle starts from the origin at t = 0 with an initial velocity of 5.3 m/s along the positive x axis.If the acceleration is (-2.6 i^ + 4.7 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.

Answer 1

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 $\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}$

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 = $-2.6m/s^{2}$

acceleration along y-axis = $4.7m/s^{2}$

Thus using the first equation of motion along x axis we get

$v_{x}=u_{x}+a_{x}t$

Applying values we get

$0=5.3-2.6t\\\\\therefore t=\frac{5.3}{2.6}sec\\\\t=2.038sec$

Now to obtain it's position we shall use third equation of motion

$v_{x}^{2}=u_{x}^{2}+2as_{x}\\\\0=(5.3)^{2}+2(-2.6)s_{x}\\\\\therefore s_{x}=\frac{-28.09}{-5.2}m\\\\s_{x}=5.402m$

Now it's location along y- axis can be obtained using 2nd equation of motion along the y axis

$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}$

Applying values as follows we get

$u_{y}=0\\a_{y}=4.7m/s^{2}\\t=2.038s$

$s_{y}=0\times 2.038+\frac{1}{2}\times 4.7m/s^{2}\times2.038^{2}\\\\s_{y}=9.76m$

thus the position vector of the particle when it reaches it's maximum x co-ordinate is

$\overrightarrow{r}=5.402\widehat{i}+9.76\widehat{j}$

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

$v_{y}=u_{y}+a_{y}t\\\\v_{y}=0+4.7\times 2.038\\\\v_{y}=9.578m/s$