Calculating the Speed of a Particle After 8.59 s

[the_d]

i have a question, it is:

A particle starts from the origin at t=o with an initial velocity having an x component of 21.8 m/s and a y component of -14.1 m/s. The particle moves in the xy plane with an x componet of acceleration only, given by 4.09 m/s^2. I need to find the speed of the particle after 8.59 s.

What I did:

i was goin to use the formula V = Vo + at but that is not correct for some reason

 

[Sirus]

You do not need to use that formula. Keep the components split; find the x and y components of the final velocity of the particle, then add them vectorally to find the speed. (Hint: you have the rate of acceleration and the time during which the particle is accelerated; how can you find the final velocity in the x direction?)

 

[Astronuc]

v = v_o + at works for the x-component of velocity.

The y component of velocity is constant.

v = v_x i + v_y j, where i is unit vector in x-direction and j is unit vector in y-direction.

or $v = \sqrt({v_x^2 + v_y^2})$

 

[jtbell]

i was goin to use the formula V = Vo + at but that is not correct for some reason

Actually, that formula is correct here. You simply need to use it as a vector formula:

$$\vec {v} = {\vec {v}}_0 + \vec {a} t$$

This corresponds to two equations, one apiece for the x and y components:

$$v_x = v_{0x} + a_x t$$

$$v_y = v_{0y} + a_y t$$

You know the components of the initial velocity and the acceleration, so you can find the components of the velocity at any time t. The speed of the particle at any time is the magnitude of the velocity vector at that time, by definition.