Question

a pilot flew a 400 - mile flight in 2.5 hours flying into the wind. flying the same rate and with the same wind speed, the return trip took only 2 hours, with a tailwind. what was the speed of the wind? 160 20 40 180

Explanation:

Step1: Define variables

Let \( p \) be the speed of the plane in still air (in miles per hour) and \( w \) be the speed of the wind (in miles per hour). When flying into the wind, the effective speed is \( p - w \). When flying with the tailwind, the effective speed is \( p + w \).

Step2: Set up equations from distance formula

The distance \( d \) is given by \( d = \text{speed} \times \text{time} \). For the trip into the wind: \( 400=(p - w)\times2.5 \). For the return trip (with tailwind): \( 400=(p + w)\times2 \).

We can simplify these equations. From the first equation: \( p - w=\frac{400}{2.5}=160 \) (Equation 1). From the second equation: \( p + w=\frac{400}{2}=200 \) (Equation 2).

Step3: Solve the system of equations

We can solve Equation 1 and Equation 2 simultaneously. Add Equation 1 and Equation 2: \((p - w)+(p + w)=160 + 200\). Simplifying the left - hand side gives \( 2p=360 \), so \( p = 180 \) miles per hour.

Now substitute \( p = 180 \) into Equation 2: \( 180+w = 200 \). Subtract 180 from both sides: \( w=200 - 180 = 20 \) miles per hour.

Answer:

20