Question

4. A particle starts at the point (8 m, 0 m) with a velocity of (-3 j) m/s. If it experiences a constant acceleration of (-3.5 i + 6 j) m/s2, after how much time does it cross the y-axis?

Answer 1

t = 1.51 s

Step-by-step explanation:

Initial position of the particle is given as

`r_1` = (8m, 0m)

now the velocity is initially along - Y direction so it is given as

`v_i = - 3\hat j`

now we know that the acceleration of the particle is given as

`\vec a = -3.5 \hat i + 6 \hat j`

now when particle will cross Y axis then the displacement in x direction is given as

`d_x = 0 - 8 = - 8m`

so in x direction we can use kinematics as

`d_x = v_(ix) t + (1)/(2) at^2`

`- 8 = 0 + (-3.5) t^2`

`t = 1.51 s`