Question

An object moving with uniform acceleration has a velocity of 13.0 cm/s in the positive x-direction when its x-coordinate is 2.76 cm. If its x-coordinate 3.05 s later is −5.00 cm, what is its acceleration?______ cm/s2

Answer 1

Since the object is moving with uniform acceleration we have an uniformly accelerated motion which means that we can use the following equations:

`\begin{gathered} a=(v_f-v_0)/(t) \\ x=x_0+v_0t+(1)/(2)at^2 \\ v_f^2-v_0^2=2a(x-x_0) \end{gathered}`

Now, in this case we know:

• The initial position 2.76 cm.

,

• The initial velocity 13 cm/s

,

• The final position -5 cm

,

• The time it takes 3.05 s.

And we want to determine the acceleration; from what we know and what we want we determine that we can use the second equation. Plugging the values in that equation we have that:

`\begin{gathered} -5=2.76+(13)(3.05)+(1)/(2)(3.05)^2a \\ (3.05^2)/(2)a=-5-2.76-(13)(3.05) \\ (3.05^2)/(2)a=-47.41 \\ 3.05^2a=-94.82 \\ a=-(94.82)/(3.05^2) \\ a=-10.19 \end{gathered}`

Therefore, the acceleration is -10.19 cm/s²