Unit 6 - Objective 3 - Application of the Integral



Position - Velocity - Acceleration

From a previous unit we know that given a position, or distance, function s(t):


Reversing the process:





Acceleration due to gravity alone is a(t) = -32 ft/sec² or -9.8 m/s². It is negative because it is acting in a downward direction.

In general, gravity is the only force affecting an item which is thrown or dropped. The equations describing it's motion would be:



If a flare is shot up from the top of an 180 foot building with a velocity of 50 feet/sec, determine each of the following.

  1. How high was it after 1/2 sec?
  2. How fast was it traveling after 1/2 sec?
  3. How high was it after 2 sec?
  4. How fast was it traveling after 2 sec?

"How high" is distance above the ground, our position function s(t).

"How fast" is the velocity function v(t). Generate the three equations of motion first.

Given, or implied in the problem are:

Integrate the acceleration to get v(t) and s(t), then use the given conditions to evaluate the constants.

Now we are ready to answer the questions.

a) s(1/2) = -16(1/2)² + 50(1/2) + 180 = -4 + 25 + 180 = 201 feet high

b) v(1/2) = -32(1/2) + 50 = -16 + 50 = 34 feet/sec


The flare is above the ground [s(t) positive], and still rising because v(t) is positive, v(t) = s'(t) is positive so s(t) is increasing.

c) s(2) = -16(2)² + 50(2) + 180 = -64 + 100 + 180 = 216 feet high

d) v(2) = -32(2) + 50 = -64 + 50 = -14 feet/sec


The flare is above ground, but now it is falling because v(t) is negative. The velocity is now directed downward. v(t) = s'(t) is negative so s(t) is now decreasing. Distance above the ground is getting smaller.

If a flare is shot up from the top of an 180 foot building with a velocity of 50 feet/sec, determine each of the following.

e) How long did it take to hit the ground?

f) How fast was it going when it hit the ground?

g) How high did it go, and at what time did it reach the highest point?



e)"How long did it take to hit the ground?" translates into: Find t when s(t) = 0. So set s(t) = 0 and solve for t.

s(t) = 16t² + 50t + 180 = 0 = -2(8t² - 25t - 90)


This is a quadratic. I don't see any factors. Using the quadratic formula gives:


the negative t value before we started.

That is not part of the solution. Discard t = -2.14. So t =5.26 sec


f) Find v(t) for the t value that makes s(t) =0

v(5.26) = -32(5.26) + 50 = -168.32 + 50 = -118.32 ft/s

Negative velocity should have been expected since it is directed downward, or because the distance above the ground is decreasing.


g)"How High" is maximum height which occurs when s'(t) = 0 = v(t).

This agrees with the fact that the flare would stop at the high point before starting downward. Set the velocity, v(t) = 0 and solve for t. We have to find at what time it reached the maximum height (critical value) before we can determine how high it is.

Now that we know when it reached the top, we can find how high that is.

The height is s(1.56) = -16(1.56)² + 50(1.56) + 180 = 219 ft



Current - Charge - Voltage


In electronics, we have the following relationships given:
q = charge measured in coulombs (C).
I = current measured in amperes (A). (rate of change of charge)
v = voltage measured in volts (V).
c = capacitance measured in farads (F).
t = time measured in seconds.




The electric current, in µA, is given by I = 1.3 - 2t. Find the amount of charge that passes a point in 0.35s





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