Question

avelocity of magnitude 40m/s is directed at an angle 40degree east of north draw a vector an paper to represent this velocity solution ​

Answer 1

Step-by-step explanation:

the horizontal and vertical component's of the velocity vector:

v_x=vcos\theta=40\ \dfrac{m}{s}\cdot cos40^{\circ}=30.6\ \dfrac{m}{s},v

x

​

=vcosθ=40

s

m

​

⋅cos40

∘

=30.6

s

m

​

,

v_y=vsin\theta=40\ \dfrac{m}{s}\cdot sin40^{\circ}=25.7\ \dfrac{m}{s}.v

y

​

=vsinθ=40

s

m

​

⋅sin40

∘

=25.7

s

m

.

Let's draw the vector (here one division corresponds to 5 m\s ):