A) one ... same current passes through each cpt
The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
A falling raindrop
Kinetic energy and potential energy are both applied when a body or object is falling.
Answer:
The train is moving with a speed of 57.6 m/s.
Explanation:
Given that,
Distance of observer from the road, d = 40 m (due south)
Velocity of the train, v = 60 m/s (due east)
So,
t is time
Net displacement is given by :
Differentiating above equation wrt t as :
Put t = 4 s
So, the train is moving with a speed of 57.6 m/s.
