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# One - x t = gt 2 v t How fast will the bullet be traveling...

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One-dimensional Motion with Constant Acceleration In the previous section on position, velocity, and acceleration we found that motion with constant acceleration is given by position functions of the form: x ( t ) = at 2 + v 0 t + x 0 where a is the acceleration (a constant), v 0 is the velocity at time t = 0 , and x 0 is the position at time t = 0 . The velocity and acceleration functions for such a position function are given by the equations v ( t ) = at + v 0 and a ( t ) = a . We will now use these equations to solve some physics problems involving motion in one dimension with constant acceleration. Firing a Bullet Directly Upwards The equation x ( t ) = - gt 2 + v 0 t + x 0 for an object moving up and down near the earth's surface can be used for more than just describing a falling object. We can also understand what happens to a bullet fired directly upwards from the surface of the earth at an initial speed v 0 . Since the initial position of the bullet is approximately x 0 = 0 , the equation for this motion is given by:
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Unformatted text preview: x ( t ) = - gt 2 + v t How fast will the bullet be traveling when it comes back down and hits the earth? To answer this we must (i) solve for the time at which the bullet will hit the earth, and (ii) find the velocity function, so that we can evaluate it at that time. Setting x ( t ) = 0 again and solving for t we find that either t = 0 or t = 2 v / g . Well, t = 0 is just the time when the bullet left the ground, so the time at which it will come back, falling from above, must be t = 2 v / g . Using our knowledge from the previous section, v ( t ) = - gt + v . If we plug in t = 2 v / g , we find that the velocity of the bullet as it comes back down and hits the ground is - g (2 v / g ) + v = - v . In other words, the bullet is traveling at the same speed it had when it was just fired, only in the opposite direction....
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