help with calculus?

One of the things you should know about position, velocity and acceleration:

f(t) = position (or height in your case)

f'(t) = velocity

f''(t) = acceleration

soooo...

a) here you should use the difference rule, so you need to find the position at the beginning and at the end and divide it by the time:

average velocity = (position at final time - position at initial time 0) / (final time - initial time)

final time = 2

initial time = 0

You should be able to find the position at the final time, being f(2), and the initial time, being f(0.)

b) To find the instintaneous velocity at any time, first you need to take the derivative of f(t), so

f'(t) = -9.8t +.... (finish the rest here)

then solve for f'(2) (velocity at t = 2 seconds)

c) To find the acceleration, you need to take the double derivative of the position formula, so

f''(t) = -9.8

Notice here that the acceleration is independent of the time, and -9.8 is actually the acceleration due to gravity on earth, which is -9.8 m/s^2. So the acceleration is always -9.8 m/s^2. The negative sign is an indication of the direction of acceleration, and if up is positive, down is negative (going towards the ground.)

d) Here, you have a situation where the ball reaches a velocity of zero for a very short moment. This is when the ball reaches its highest point. So what you need to do here is solve for the time it takes to get there by substituting f'(t) = 0 (velocity = 0) in the equation you found in b. Solve for t, then substitute that answer "t" back into the position formula f(t) to determine the actual height.

e) The length of time it spends in the air is twice the time it takes to reach the top.. I think.

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Someone please correct me if I'm wrong.. I haven't taken a calculus course in a few years. I would hate to misguide our troubled youth. lol..

Equation: x = x(i) + v(i)t + (at^2)/2, to find position once you know a (acceleration), v(i) (initial velocity) and x(i) (initial position).

From your data: x(i) = 3 m, v(i) = 25 m/s, a = -9.8 m/s^2

a) Position after 2 secs = -4.9 x 4 + 50 + 3 = 33.4 m

Distance traveled in 2 secs: 33.4 - 3 = 30.4 m

Average velocity = 30.4 m / 2 secs = 15.2 m/s

b) Derive for t: v(t) = v(i) + at = 25 - 9.8t, for 2 seconds = 5.4 m/s

c) Derive for t again: a = a, which is constant = -9.8 m/s^2

d) Highest point is when v(t) = 0, that is: 0 = v(i) + at --> -at = v(i)

9.8t = 25, then t = 2.6 s

Replace in the equation: - 4.9 (2.6)^2 + (25 x 2.6) + 3 = -33 + 65 + 3 = 35 meters high.

e) find t when x = 0:

0 = -4.9t^2 + 25t + 3, just solve the quadratic equation and take the positive value.

a)

Average velocity = change in speed / change in time

=f(2) - f(0) / 2 - 0

=33.4 - 3 / 2

=15.2 m/sec

b)

Instantaneous velocity = derivative of position function:

v(t) = f'(t) = -9.8t + 25

f'(2) = 5.4 m/s

c)

Acceleration = derivative of velocity function:

f''(t) = -9.8 m/sec^2

d)

At the maximum height, the ball stops and starts falling back to earth, so its velocity is zero at this point:

v = 0 = -9.8t + 25

t = 2.6 sec

At 2.6 seconds, f(2.6) = f(t) = - 4.9(2.6)^2 + 25(2.6) + 3

f= 34.9 meters

e)

When the baseball starts the position is zero, and when the baseball returns its position is also zero. Set f(t) = 0 and solve for "t" using quadratic formula:

f(t) = -4.9 t ^ 2 + 25 t + 3 = 0

t = {-25 +/- sqrt[ (25^2) - (4*-4.9* 3) ] } / (2* -4.9)

t=

5.2 sec

-0.1 sec

Since t cannot be negative, the ball is in the air for 5.2 seconds.

Note also that 5.2 seconds is twice the time it hit its maximum height, but that was half the ball's journey.

It looks like you want answers not help. Write down where you are stuck and perhaps I'll help. There are simple equations in the book. Why not try them and perhaps you'll learn.