Question

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane travels 1.11 m. How much additional distance will the airplane travel during the second second of its motion?

Answer 1

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$.

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider  the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$, then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$.

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$