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 answerâ†“

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

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

s_1=20 ft

#### Final Answer:

Position of particle = 20 ft