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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?

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

Respuesta :

Edufirst
Data:

1) Vi = 58.0 m/s

2) Vf = 153 m/s

3) t = 12.0 s

4) D = ?

Formulas:

Uniform acceleration =>

a = (Vf - Vi) / t

d = Vi + a* t^2 / 2

Solution

a = ( 153 m/s - 58.0 m/s) / (12.0 s) = 7.92 m/s^2

d = 58.0 m/s + 7.92 m/s^2 * (12.0s)^2 / 2 = 628 m

628 m = 6.28 * 10^2 m

Answer: option E) 6.28 * 10^2 meters

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