Question

1. an engineer must design a runway to accommodate airplanes that must reach a ground speed of 61 m/s before they can take off. these planes are capable of being accelerated uniformly at the rate of 2.5 m/s².

a. how long will it take the planes to reach take - off speed?
b. what must the minimum length of the runway be?

Explanation:

Part (a)

Step1: Identify known quantities

We know the initial velocity \( v_i = 0\) m/s (starts from rest), final velocity \( v_f=61\) m/s, and acceleration \( a = 2.5\) m/s². We use the kinematic equation \( v_f=v_i + at\).

Step2: Solve for time \( t\)

Rearrange the equation \( v_f=v_i + at\) to solve for \( t\): \( t=\frac{v_f - v_i}{a}\). Substitute the values: \( v_i = 0\), \( v_f = 61\) m/s, \( a=2.5\) m/s². So \( t=\frac{61 - 0}{2.5}\).

Step3: Calculate the time

\( t=\frac{61}{2.5}=24.4\) s.

Part (b)

Step1: Choose a kinematic equation

We can use the equation \( v_f^2=v_i^2 + 2ad\) to find the distance \( d\) (length of runway). We know \( v_i = 0\), \( v_f = 61\) m/s, \( a = 2.5\) m/s².

Step2: Solve for \( d\)

Rearrange the equation \( v_f^2=v_i^2+2ad\) to \( d=\frac{v_f^2 - v_i^2}{2a}\). Substitute \( v_i = 0\), \( v_f = 61\) m/s, \( a = 2.5\) m/s². So \( d=\frac{61^2-0^2}{2\times2.5}\).

Step3: Calculate the distance

First, calculate \( 61^2 = 3721\). Then \( d=\frac{3721}{5}=744.2\) m.

Answer:

s:
a. \(\boldsymbol{24.4}\) seconds
b. \(\boldsymbol{744.2}\) meters