Question

A jet airliner moving initially at 366 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 127 mph in a direction 60◦ north of east. What is the new speed of the aircraft with respect to the ground?

Answer 1

Answer:

The new speed is $V_r = 443.4 \ m/s$ , in the direction  $\theta = 14.36^o$ with respect to the ground

Explanation:

From the question we are told that

   The initial speed of the jet airliner is 366 mph to the east (positive x - axis )

    The direction of the wind is  $u = 127 \ mps$ toward  60° north west

     Generally the  x component of jet airliner velocity is  

             $v_x = 366 * cos (0) = 366 \ mph$

      Generally the  y component of jet airliner velocity is

              $v_y = 366 * sin (0) = 0 \ mph$

       Generally the  x component of winds velocity is  

              $u_x = 127 cos(60) = 63.5 \ m/s$

        Generally the y component of winds velocity is

              $u_x = 127 sin(60) = 109.99 \ m/s$

Generally the velocity of both the jet airliner and wind along the x-axis is

        $V_x = v_x + u_x = 366 + 63.5 = 429.5 \ m/s$

Generally the velocity of both the jet airliner and wind along the y-axis is

        $V_y = v_y + u_y = 0 + 109.99 = 109.99 \ m/s$

Generally the resultant velocity is  

          $V_r = \sqrt{V_x^2 + V_y^2}$

=>       $V_r = \sqrt{429.5^2 + 109.99^2}$

=>       $V_r = 443.4 \ m/s$

 The direction is  

         $\theta = tan^{-1} [\frac{V_y}{V_x}]$

=>      $\theta = tan ^{-1}[\frac{109.99}{429.5}]$

=>    $\theta = 14.36^o$