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A spaceship starting from a resting position accelerates at a constant rate of 9.8 m/s. How far will the spaceship travel if its final speed is 1 percent of the speed of light (300, 000, 000 m/s?)

Respuesta :

osprey2015

300 000 0 squared = 2 x 9.8 distance

KINEMATICS

Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.


speed = distance/time

velocity = displacement/time

s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time


v=u+at

v^2=u^2+2as

s=ut+1/2at^2

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