# An airplane accelerates at "3.20 m/s"^2" over "32.8 s" before takeoff. How far does it travel before takeoff?

Oct 12, 2016

The airplane travels 1720 m before takeoff.

#### Explanation:

Based on the variables in your question, the following equation will be used.

$d = {v}_{i} t + \frac{1}{2} a {t}^{2}$, where $d$ is distance, ${v}_{i}$ is initial velocity, $t$ is time, and $a$ is acceleration.

Known Values
${v}_{i} = 0$
$a = \text{3.20 m/s"^2}$
$t = \text{32.8 s}$

Unknown Value
$d$

Solution
Since ${v}_{i} = 0$, ${v}_{i} t = 0$. Therefore we can eliminate ${v}_{i} t$ from the equation.

Substitute the known values into the equation and solve.

$d = \frac{1}{2} a {t}^{2}$

$d = \frac{a {t}^{2}}{2}$

$d = \frac{{\left(3.20 \text{m"/"s"^2)*(32.8"s}\right)}^{2}}{2}$

d=(3.20"m"/cancel("s"^2)*1075.84cancel("s"^2))/2="1720 m" rounded to three significant figures