Question

An airplane starts at rest on a runway. It accelerates at `4.44 m/s^2`, and travels 738 m before taking off. How long did it take?

Answer 1

Reverse-engineer, using calculus, the definition of acceleration (then velocity) to solve for displacement as a function of time, then equate that to the given displacement to solve for time, yielding

`t = (30sqrt(4551))/111 s ~~ 18.23272970326803 s~~18.2s`

Explanation:

Let's see what we have:

`a = 4.44 m/s^2`

`s = 738 m`

And we're trying to find the time it took for an object to travel `738m` with a constant acceleration of `4.44 m/s^2` starting from rest.

Here's a tactic we can try: since acceleration relates to both time and displacement, we could pretend to be solving for displacement from acceleration, which will yield a function of time, then equate our displacement function with the given displacement, to solve for time.

To solve for displacement from acceleration, we could first solve for velocity from acceleration. In fact, we can do this by looking at the calculus definition of acceleration:

`a = (dv)/(dt)`

Integrating both sides to solve for velocity:

`int a dt = int (dv)/(dt) dt`

`int a dt = v`

Let's plug in our acceleration:

`v = int (4.44 m/s^2) dt`

Pull the scalar out:

`v = 4.44 m/s^2 int (1) dt`

And solve:

`v = 4.44 m/s^2 * (t + C)`

The constant of integration (`C`) in this context refers to the starting velocity, but since the starting velocity is zero (because the object starts at rest), we can ignore the constant:

`v = 4.44t m/s^2`

Don't worry about the units, as the time variable will have a unit `s`, canceling out with `m/s^2`, becoming `m/s`.

Now that we have velocity, we could use the definition of velocity, which relates to displacement:

`v = (ds)/(dt)`

Take the integral with respect to time:

`int v dt = int (ds)/(dt) dt`

`int v dt = s`

And solve for displacement, plugging in the value for velocity that we have solved earlier:

`s = int (4.44t m/s^2) dt`

Taking out the scalar once again:

`s = 4.44 m/s^2 int (t) dt`

And solving:

`s = 4.44 m/s^2 * (t^2/2 + C)`

Again, the constant of integration doesn't matter here:

`s = 2.22t^2 m/s^2`

Now, we can equate this with the given displacement:

`2.22t^2 m/s^2 = 738m`

We might be able to divide by `2.22 m/s^2`:

`(2.22t^2 m/s^2)/(2.22 m/s^2) = (738m)/(2.22 m/s^2)`

Simplifying:

`t^2 = 738 m * 1/(2.22 m/s^2)`

`t^2 = 738 m * 1/2.22 s^2/m`

`t^2 = 738 * 1/2.22 s^2`

`t^2 = 738 * 1/(222/100) s^2`

`t^2 = 738 * 1/(111/50) s^2`

`t^2 = 738 * 50/111 s^2`

`t^2 = (738 * 50)/111 s^2`

`t^2 = 36900/111 s^2`

Taking the square root:

`sqrt(t^2) = sqrt(36900/111 s^2)`

`t = sqrt(36900/111) s`

And simplifying further:

`t = sqrt(36900)/sqrt(111) s`

`t = sqrt(41 * 100 * 9)/sqrt(111) s`

`t = (sqrt(900)sqrt(41))/sqrt(111) s`

`t = (30sqrt(41))/sqrt(111) s`

`t = (30sqrt(41))/sqrt(111) * sqrt(111)/sqrt(111) s`

`t = (30sqrt(41)sqrt(111))/111 s`

`t = (30sqrt(4551))/111 s`

Plugging that into a calculator, we get approximately...

`(30sqrt(4551))/111 ~~ 18.23272970326803`

Therefore, the time it takes for an object with a constant acceleration of `4.44 m/s^2` to be displaced `738m` starting at rest is

`t = (30sqrt(4551))/111 s ~~ 18.23272970326803 s~~18.2s`

Answer 2

The time is `=7.32s`

Explanation:

A very simple way of tackling this problem.

Apply the equation of motion

`s=ut+1/2at^2`

The initial velocity is `u=0ms^-1`

The acceleration is `a=4.44ms^-2`

The distance is `s=738m`

Therefore,

`738=0xxt+1/2xx4.44xxt^2`

`t^2=(2*738)/4.44`

`t=sqrt((2*738)/4.44)=18.2s`