Question

A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?

Answer 1

accelaration a:
a = (v2 - v1)/t
distance s:
s = (v2^2 - v1^2)/(2*a) = 12*(v2 + v1)/2 = 6 s *(153 m/s + 58 m/s) = 1.27*10^3 m