Question

A particle leaves the origin with an initial velocity of 4.90 m/s in the x direction, and moves with constant acceleration ax = -1.50 m/s^2 and ay = 3.00 m/s^2. How far does the particle move in the x direction before turning around?

Answer 1

Answer:

it will move by d = 8.00 m in x direction before it will turn back

Explanation:

Here the initial velocity of particle is along +x direction

it is given as

$v_i = 4.90 m/s$

now its acceleration is given as

$a_x = -1.50 m/s^2$

$a_y = 3.00 m/s^2$

now when it turns back then the velocity in x direction will become zero

so we will say

$v_f^2 - v_i^2 = 2 a d$

$0 - 4.90^2 = 2(-1.50) d$

$d = 8.00 m$

so it will move by d = 8.00 m in x direction before it will turn back