# If a particle has an initial velocity

If a particle has an initial velocity of $v_0 = 12$ ft/s to the right, at $s_0=0$, determine its position when $t = 10$ s, if $a = 2$ ft/$s^2$ to the left.

#### Solution:

Show me the final answer↓

To find the position of the particle, we will use the following formula:

$s_1=s_0+v_0t+\dfrac{1}{2}a_ct^2$

(Where $s_1$ is final displacement, $s_0$ is initial displacement, $v_0$ is initial velocity, $t$ is time, and $a_c$ is constant acceleration)

The following are given to us in the question:

$v_0=12$ ft/s

$s_0=0$ ft

$t=10$ s

$a=-2$ ft/$s^2$

(note that our acceleration is negative because we are assuming objects moving to the right to be positive. The particle is moving to the right, however the acceleration is to the left, thus, a negative acceleration)

Substitute the values into our equation:

$s_1=s_0+v_0t+\dfrac{1}{2}a_ct^2$

$s_1=0+(12)(10)+\dfrac{1}{2}(-2)(10^2)$

$s_1=20$ ft

#### Final Answer:

Position of particle = 20 ft