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

Answer on Question #51423, Physics, Astronomy Astrophysics

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?

Solution:

Change of speed is described by the Eq.(1)

$$v = v _ {0} + a t$$

where $v_{0} = 58m / s$ is the initial velocity; $a$ is the acceleration.

From Eq.(1)

$$a = \frac {v - v _ {0}}{t}$$

The change of coordinates is described by the equation

$$x = v _ {0} t + \frac {a t ^ {2}}{2}$$

From Eq.(2) and Eq.(3)

$$x = v _ {0} t + \frac {\left(v - v _ {0}\right) t ^ {2}}{2 t} = t \left(v _ {0} + \frac {v - v _ {0}}{2}\right) = \frac {v + v _ {0}}{2} t = \frac {5 8 + 1 5 3}{2} \cdot 1 2 = 1 2 6 6 m$$

Answer: 1266m

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