Aparticle 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.

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

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 =

acceleration along y-axis =

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

Applying values we get

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

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

Applying values as follows we get

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

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

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