<p>A ball is thrown vertically upward from the ground with an initial velocity of 32ft/s. Let t seconds be the time that has elapsed since the ball was thrown and s feet be the distance of ball from the starting point at t seconds.
a.What is the position function.
b.Acceleration function.
c.Velocity function.
d.acceleration at t=1.5 s.</p>
<p>My answers( I am not sure if these are right)
a.-16t^2 + 32t
b.-32t+32
c.I've no clue
d.I can find that out once I know answer to c.</p>
<p>I think you got B and C mixed up. </p>
<p>Velocity is the derivative of s(t) (position), while acceleration is double prime of s(t). So, acceleration would be the derivative of velocity. </p>
<p>I'm also confused about why you made the velocity negative when it would be a positive slope, am I missing something?</p>
<p>acceleration due to gravity is 32 ft/s^2, but in this case the acceleration is negative because the ball is thrown upward while the gravity pulls it downward. The acceleration is also constant. so, the acceleration function is a(t) = -32. Velocity function is the intergral (antiderivative) of the acceleration function, so v(t) = -32t + C. C represents the initial velocity, which is 32. So v(t) = -32t + 32. Antiderivative of the v(t) would yield the position function, s(t) = -16t^2 + 32t + C. C, the initial position, is 0 because the ball is thrown from the ground. So, s(t) = -16t^2 + 32t.</p>
<p>So your answers to the problem will be
a. position function s(t) = -16t^2 + 32t (ur answer was correct)
b. acceleration function a(t) = -32 (what u wrote down was the velocity function)
c. velocity fuction v(t) = -32t + 32
d. acceleration at t=1.5s a= -32 ft/s^2, since acceleration is constant.</p>
<p>I hope I'm right...</p>
<p>^^</p>
<p>Alright, sweet.</p>