Question

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is (-3.2 i^ + 4.6 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate.

Answer 1

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Step-by-step explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

`v_x = 4.8 m/s`

now the acceleration is given as

`\vec a = -3.2 \hat i + 4.6 \hat j`

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

`v_f = v_i + at`

`0 = 4.8 - 3.2 t`

`t = 1.5 s`

Part a)

the velocity of the particle at this moment in Y direction is given as

`v_f_y = v_i + at`

`v_f_y = 0 + 4.6(1.5)`

`v_f_y = 6.9 m/s`

Part b)

X coordinate of the particle at this time

`x = v_x t + (1)/(2)a_x t^2`

`x = 4.8(1.5) - (1)/(2)(3.2)(1.5^2)`

`x = 3.6 m`

Y coordinate of the particle at this time

`y = v_y t + (1)/(2)a_y t^2`

`y = 0(1.5) + (1)/(2)(4.6)(1.5^2)`

`y = +5.175 m`

so position is given as (3.6 m, 5.175 m)