# If an object with uniform acceleration (or deceleration) has a speed of 5 m/s at t=0 and moves a total of 15 m by t=12, what was the object's rate of acceleration?

Mar 30, 2016

Any acceleration problem can be solved given three variables and the kinematics equations.

#### Explanation:

In kinematics problems, there are five variables:
${\vec{v}}_{i} \text{ }$ (initial velocity)
${\vec{v}}_{f} \text{ }$ (final velocity)
$\vec{a} \text{ }$ (acceleration)
$\Delta \vec{d} \text{ }$ (displacement)
$\Delta t \text{ }$ (elapsed time)

and five equations, each of which only uses four of the variables:

$\Delta \vec{d} = {\vec{v}}_{i} t + \frac{1}{2} \vec{a} \Delta {t}^{2}$
$\Delta \vec{d} = {\vec{v}}_{f} t - \frac{1}{2} \vec{a} \Delta {t}^{2}$
$\Delta \vec{d} = \frac{{\vec{v}}_{i} + {\vec{v}}_{f}}{2} \Delta t$
$\vec{a} = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{\Delta t}$
${\vec{v}}_{f}^{2} = {\vec{v}}_{i} 2 + 2 \vec{a} \Delta \vec{d}$

so to solve any question, we need three variables, and the equation that has the fourth we are trying to find. In this question, we have our initial velocity, displacement, and time, and we are looking for acceleration. The equation that has those four variables is:

$\Delta \vec{d} = {\vec{v}}_{i} t + \frac{1}{2} \vec{a} \Delta {t}^{2}$

so let's insert our values and solve for the unknown:

$15 m = 5 \frac{m}{\cancel{s}} \cdot 12 \cancel{s} + \frac{1}{2} \vec{a} {\left(12 s\right)}^{2}$
$15 m = 60 m + \frac{1}{2} \cdot 144 {s}^{2} \cdot \vec{a}$
$15 m - 60 m = 72 {s}^{2} \cdot \vec{a}$
$\frac{- 45 m}{72 {s}^{2}} = \vec{a}$
$\vec{a} = - 0.625 \frac{m}{s} ^ 2$

This process works for any acceleration problem where we have three of the five variables, and are looking for a fourth.