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Hi there.
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So for this problem, we are told that a projectile is shot upward from the surface of earth with an initial velocity.
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That initial velocity we call it b -0 and that is equal to 111 meters per second.
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And so we need to use the position function that is given.
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That position function is the following.
00:28
Is x of t is equal to 4 .9 times the time to the squared plus the initial speed times the time plus the initial position.
00:47
So in this case we are asked what is the velocity after two seconds and after when the time is 2 seconds and when the time is 14 seconds.
01:04
So in here, what we need first to do is to derivate this expression because we know that the derivative of the position function corresponds to the velocity function.
01:19
So we know that the velocity function that depends on the time is equal to the derivative of the position function, respect to time.
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So from here we obtained that this is 2 times 4 .9 meters per second square.
01:36
And this times the time plus the initial speed...