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

The answer is 3.33m

Step-by-step explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

`(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.

`(dx)/(dt) = at \rightarrow \int_(x_0)^x dx = a\int_0^t t\ dt`

then

`x- x_0 = (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 = (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`