A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is
1 answer:
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Answer
given,
v = (6 t - 3 t²) m/s
we know,
position of the particle
integrating both side
x = 3 t² - t³
Position of the particle at t= 3 s
x = 3 x 3² - 3³
x = 0 m
now, particle’s deceleration
a = 6 - 6 t
at t= 3 s
a = 6 - 6 x 3
a = -12 m/s²
distance traveled by the particle
x = 3 t² - t³
at t = 0 x = 0
t = 1 s , x = 3 (1)² - 1³ = 2 m
t = 2 s , x = 3(2)² - 2³ = 4 m
t = 3 s , x = 0 m
total distance traveled by the particle
D = distance in 0-1 s + distance in 1 -2 s + distance in 2 -3 s
D = 2 + 4 + 2 = 8 m
average speed of the particle
To solve this problem it is necessary to apply the concepts related to the orbital velocity of a satellite on earth.
This concept is expressed in the equation,
Where,
G = Universal Gravitational constant
Mass of the Earth
Therefore the ratio of the velocity from two satellites is,
The ratio between the two satellites is the same, then
Therefore the correct option is B.