Question #10898

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?

Expert's answer

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

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

## Comments

Assignment Expert24.06.15, 19:14Dear bianca, the rigtht answer is B.

bianca19.06.15, 17:29which one is is it i still dont get it?

A. 6.96 × 102 meters

B. 1.27 × 103 meters

C. 5.70 × 102 meters

D. 1.26 × 102 meters

E. 6.28 × 102 meters

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