a velocity of magnitude 40m/s is directed at an angle of 40degree east of north draw a vector on paper to represent this velocity
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Answer:
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 ):
Explanation:
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