An object moving with uniform acceleration has a velocity of $11\mathrm{.}5c\frac{m}{s}$ in the positive $x-$direction when its $x-$

coordinate is $2\mathrm{.}90cm.$ If its $x-$coordinate $2\mathrm{.}40s$ later is $-5\mathrm{.}00cm,$ what is its acceleration?

Step $1$

The object has moved to a particular coordinate in the positive $x-$direction with a certain velocity and constant

acceleration; then it reverses its direction and moves in the negative $x-$direction to a particular $x-$coordinate in

time $t.$ We are given an initial velocity $v_i=11\mathrm{.}5c\frac{m}{s}$ in the positive $x-$direction when the initial position is

$x_i=2\mathrm{.}90cm(t=0).$ We are given that at $t=2\mathrm{.}40s,$ the final position is $x_f=-5\mathrm{.}00cm.$ The acceleration is

uniform so that we have the following equation in terms of the constant acceleration $a.$

$x_f-x_i=v_{i}t+\frac{1}{2}a{t}^2$

Now we substitute the given values into this equation.

$2\mathrm{.}4s$

Step $2$

Simplifying the equation from the previous step further, we obtain the following.

$(cm)=(cm)+(s^2)a$

Solving for a gives

$a=$

Your response differs from the correct answer by more than $10\%.$ Double check your calculations. $c\frac{m}{s^2}.$