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

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

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

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

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