Question

A fighter jet travels 10km at an angle of 30 degrees. He then turns sharply and flies 25 km at an angle of 75 degrees. What is the fighter jets displacement

Answer 1

Answer:

24.4 km

Explanation:

First displacement , d' = 10 km at an angle 30°

second displacement, d'' = 25 km at an angle 75° with the previous one

Write the displacements in the vector form

$\overrightarrow{d'} =10 ( Cos 30 \widehat{i} + Sin 30 \widehat{j})$

$\overrightarrow{d'} =8.66\widehat{i} + 5\widehat{j}$

$\overrightarrow{d''} =25 ( - Cos 45 \widehat{i} + Sin 45 \widehat{j})$

$\overrightarrow{d''} =-17.68\widehat{i} + 17.68\widehat{j}$

The net displacement is given by

$\overrightarrow{d}= \overrightarrow{d'}+ \overrightarrow{d''}$

$\overrightarrow{d}= - 9.02 \widehat{i} + 22.68 \widehat{j}$

the magnitude of the displacement is given by

$d=\sqrt{9.02^{2}+22.68^{2}}$

d = 24.4 km