Question

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. What is the speed of the airplane at the end of the second second?

Answer 1

To find the speed of the airplane at the end of the second second, we can use the equation for constant acceleration and calculate the acceleration based on the distance traveled in the first second. By substituting the values into the formula, we can find the final velocity.

Step-by-step explanation:

To find the speed of the airplane at the end of the second second, we need to use the equation for constant acceleration.

The formula to find the final velocity is:

v = u + at

Where:

v is the final velocity,

u is the initial velocity (which is 0 in this case since the airplane starts from rest),

a is the acceleration, and

t is the time.

In the first second, the airplane travels 1.11 m. Since we know the initial velocity (0) and the distance traveled (1.11 m), we can rearrange the formula to solve for acceleration:

a = (v - u) / t

Substituting the given values:

a = (1.11 - 0) / 1

a = 1.11 m/s²

Now we can use the same formula to find the final velocity at the end of the second second:

v = u + at

v = 0 + (1.11 * 2)

v = 2.22 m/s

Therefore, the speed of the airplane at the end of the second second is 2.22 m/s.

Answer 2

`v=4.44(m)/(s)`

Step-by-step explanation:

Given that the airplane starts from the rest (this is initial velocity equals to zero) and accelerates at a constant rate, position can be described like this:
`x=v_(0)t +(1)/(2) at^(2)` where x is the position, t is the time a is the acceleration and
`v_(0)` is initial velocity. In this way acceleration can be found.
`a=(2(x-v_(0)t) )/(t^(2) ) =(2(1.11m-0))/(1s^(2) ) =2.22(m)/(s^(2) )`.

Now we are able to found velocity at any time with the formula:
`v=v_(0) +at = 0(m)/(s) +(2.22(m)/(s^(2)).2s)=4.44(m)/(s)`