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

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

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Answer 1 · 2016-10-12T21:15:36

The airplane travels 1720 m before takeoff.

Explanation:

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

$d = v_{i} t + \frac{1}{2} a t^{2}$ $d=v_it+1/2at^2$ , where $d$ $d$ is distance, $v_{i}$ $v_i$ is initial velocity, $t$ $t$ is time, and $a$ $a$ is acceleration.

Known Values
$v_{i} = 0$ $v_i=0$
$a = {3.20 m/s}^{2}$ $a="3.20 m/s"^2"$
$t = 32.8 s$ $t="32.8 s"$

Unknown Value
$d$ $d$

Solution
Since $v_{i} = 0$ $v_i=0$ , $v_{i} t = 0$ $v_it=0$ . Therefore we can eliminate $v_{i} t$ $v_it$ from the equation.

Substitute the known values into the equation and solve.

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

$d = \frac{a t^{2}}{2}$ $d=(at^2)/2$

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

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

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

1 Answer

Explanation:

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1 Answer

Explanation:

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