Question

A car travels 6 km towards north at an angle of 45 to the east and then travels distance of 4 km towards north at an angle of 135 to the east. how far is the point from the starting point? what angle does the straight line joining its instial and final position makes with the east?

Answer 1

HELLO DEAR,

now, net distance along x-direction.

$\bold{S_x = 6cos45 - 4cos45}$

$\bold{S_x = 2cos45}$

$\bold{S_x = 2*1/\sqrt{2} = \sqrt{2}km}$

net distance along y-direction

$\bold{S_y = 6sin45 + 4sin45}$

$\bold{S_y = 10sin45}$

$\bold{S_x = 10*1/\sqrt{2} = 5\sqrt{2}km}$

$\bold{\therefore}$ net distance travelled from the starting point ,

$\bold{\sqrt{(S_x)^2 + (S_y)^2}}$

$\bold{\sqrt{(\sqrt{2})^2 + (5\sqrt{2})^2}}$

$\bold{\sqrt{2 + 50}}$

$\bold{\sqrt{52}}$

now,the angle which resultant make with east direction

$\bold{tan\theta = \frac{y-COMPONENT}{x-COMPONENT}}$

$\bold{\tan\theta = \frac{5\sqrt{2}}{\sqrt{2}} = 5}$

$\bold{\theta = tan^{-1}5}$

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Answer:

just see the above answer

Step-by-step explanation:

it is better than my answer