Velocity and acceleration

Chapter 20 - 4 Velocity and Acceleration

Derivatives can be related very easily to physics applications. We can relate it to the position function, usually denoted as s(t) or h(t), the velocity function denoted v(t), and the acceleration function denoted a(t). Notice that are a function of time! Warning!! Warning!! Make sure you understand the difference between average and instantaneous. The average velocity can be described as the change between two points, thus giving you the slope of the line connecting these two points! While the instantaneous velocity gives you the slope at a single point in time, thus giving you the slope of the tangent line! That's right!! It's the derivative!!!

Average formulas

Where "s" is the position at any time "t"

Instantaneous Formulas

v(t) = s'(t) a(t) = v'(t) = s"(t)

Where v(t) is the first derivative of the position function and a(t) is the first derivative of the velocity function. Also note it is the second derivative of the position function!!

Comments v(t) = 0 means the object is not moving. v(t) = 3 means the object would be moving forward (positive) v(t) = -2 means the object would be moving backward (negative) a(t) = 0 means there is no change in the velocity a(t) = 5 means the object is going faster (positive) a(t) = -3 means the object is slowing down.

Sample problems

1) s(t) = t² - 20. find the average velocity from t = 3 to t = 5.

Solution: Use the average velocity formula!

Therefore, the average velocity is 8

2) Find the velocity function and the acceleration function for the function s(t) = 2t³ + 5t - 7

Solution: Use the instantaneous formulas

v(t) = s'(t) = 6t² + 5 a(t) = v'(t) = 12t

b) Find the velocity and acceleration at t = 2 for the above function.

Solution: Simply replace t with 2 in the above formulas

v(2) = 6(4) + 5 = 29 a(2) = 12(2) = 24

3) If a ball is thrown vertically upward with an initial velocity of 128 ft/sec, the ball's height after t seconds is s(t) = 128t - 16t²

a) What is the velocity function?

Solution: v(t) = 128 - 32t

b) What is the velocity when t = 2, 4, 6

Solution: v(2) = 128 - 32(2) = 64

v(4) = 128 - 32(4) = 32

v(6) = 128 - 32(6) = -64

c) At what time is the velocity 48 ft/sec? 16 ft/sec? -48 ft/sec?

Solution: Set the velocity function to each of the above values!

48 = 128 - 32t 16 = 128 - 32t -48 = 128 - 32t -80 = -32t -112 = -32t -176 = -32t

2.5 = t 3.5 = t 5.5 = t

d) When is the velocity zero?

Solution: Set the velocity function to zero and solve

0 = 128 - 32t -128 = -32t 4 = t

e) What is the height of the ball at the time the velocity is 0?

Solution: let t = 4 in the position function

s(4) = 128(4) - 16(16) = 512 - 256 = 256'

f) Is the answer to part "e" the maximum height?

Solution: Yes, because it is the zero of the first derivative.

g) When does the ball hit the ground?

Solution: Set the position function to zero and solve!

128t - 16t² = 0 16t(8 - t) = 0 At t = 0, or t = 8. Think. At t = 0, you haven't thrown the ball yet! The ball will hit the ground in 8 seconds. You could have reasoned that it took 4 seconds to reach maximum height and another 4 secs to come back down.

h) What is the velocity when the ball hits the ground?

Solution: Find v(8) v(8) = 128 - 32(8) = -128 ft/sec. It is negative because the ball is coming back down! Notice it has the same velocity as when it was thrown upward!

i) What is the acceleration function?

Solution: Find a(t) by finding the derivative of v(t)!

a(t) = -32 ft/sec²

j) What is the acceleration at t = 3, 5, 8?

Solution: a(3) = -32 a(5) = -32 a(8) = -32

No matter what time it will always be -32. (This is the effect of gravity.)

It's now time to get ready for the big test!! Sample test is next so have plenty of water (notice the motif) and hang on to your hat!