Question

An object moving with uniform acceleration has a velocity of $12.0cm/s$ in the positive $x$ direction when its $x$ coordinate is $3.00cm$. If its $x$ coordinate $2.00s$ later is $25.00cm$, what is its acceleration?

Answer 1

The velocity is always changing; there is always nonzero acceleration and the problem says it is constant. So we can use one of the set of equations describing constant-acceleration motion. Take the initial point to be the moment when $x_i=3.00cm$ and $v_{xi}=12.0cm/s$. Also, at

$t=2.00s,x_f=–5.00cm$.

Once you have classified the object as a particle moving with constant acceleration and have the standard set of four equations in front of you, how do you choose which equation to use? Make a list of all of the six symbols in the equations: $x_i,x_f,v_{xi},v_{xf},a_x$, and $t$. On the list fill in values as above, showing that $x_i,x_f,v_{xi}$, and $t$ are known. Identify $a_x$ as the unknown. Choose an equation involving only one unknown and

the knowns. That is, choose an equation not involving $v_{xf}$. Thus we choose the kinematic equation

$x_f=x_i+v_{xi}+\frac{1}{2}{a}_x{t}^2$

and solve for $a_x$

$a_x=\frac{2[x_f-x_i-v_{xi}t]}{t^2}$

We substitute,

$a_x=\frac{2[-5.00cm-3.00cm-(12.0cm/s\left)\right(2.00s\left)\right]}{(2.00s{)}^2}=-16.0cm/s^2$