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Calculus AB: Applications of the Derivative

Rates of Change and Applications to Motion


Problems for "Rates of Change and Applications to Motion"

Average Rates of Change

Suppose s(t) = 2t 3 represents the position of a race car along a straight track, measured in feet from the starting line at time t seconds. What is the average rate of change of s(t) from t = 2 to t = 3 ?

The average rate of change is equal to the total change in position divided by the total change in time:

Avg Rate   =  
    = 38 ft per second  

In physics, velocity is the rate of change of position. Thus, 38 feet per second is the average velocity of the car between times t = 2 and t = 3 .

Instantaneous Rates of Change

What is the instantaneous rate of change of the same race car at time t = 2 ?

The instantaneous rate of change measures the rate of change, or slope, of a curve at a certain instant. Thus, the instantaneous rate of change is given by the derivative. In this case, the instantaneous rate is s'(2) .

s'(t) = 6t 2  
s'(2) = 6(2)2 = 24 feet per second  

Thus, the derivative shows that the racecar had an instantaneous velocity of 24 feet per second at time t = 2 .

Rectilinear Motion

The kind of motion that was discussed above is called rectilinear motion, which refers to the motion of an object in a straight line. Such motion can be depicted as a point which moves forwards and/or backwards on a number line.

General Motion Equations

If s(t) represents the position of the object on the number line at time t , then v(t) , the (instantaneous) velocity, is equal to s'(t) , and a(t) , the (instantaneous) acceleration, is equal to v'(t) , which is s''(t) .

Thus, velocity is the rate of change of position, and acceleration is the rate of change of velocity.


If s(t) = t 2 - 5t , what is the position, velocity and acceleration at t = 2 ? Assume s is in feet and t is in seconds, and interpret these results.

s(t) = t 2 - 5t + 3
v(t) = s'(t) = 2t - 5
a(t) = v'(t) = 2

s(2) = 2
v(2) = - 1
a(2) = 2

So, at t = 2 , the object is located at +2 feet from the start. The velocity is -1 foot per second. The negative sign indicates that it is headed in the negative direction, and it is moving backwards at a rate of one foot per second. The acceleration is 2, which means that at that instant, its velocity is increasing by a rate of 2 feet per second each second.

Vector and Scalar Quantities

Position, velocity, and acceleration are all vector quantities because they contain both a direction and a magnitude. For example, if the velocity of an object is -3 feet per second, then that object is moving backwards (direction) at a rate of 3 feet per second (magnitude). Similarly, if an object has a position of -3 feet, then is located 3 feet from the starting point (magnitude), but on the negative side (direction).

The vector quantities of position and velocity both have corresponding scalar quantities that only have a magnitude. The scalar analog of position is distance. Although the position of an object with respect to the start line may be -3 feet, its distance from that start line is simply 3 feet, because distance is always a positive quantity. Thus, distance is the absolute value of position.

Similarly, the scalar analog of velocity is speed. Whether an object's velocity is -5 feet per second, or +5 feet per second, its speed is still simply 5 feet per second, because speed is always a positive quantity that contains no information about direction. Thus, speed is the absolute value of velocity.

Motion With Constant Acceleration

This section refers to the special case of rectilinear motion in which the acceleration is constant. In cases where the acceleration is constant, a(t) can be represented simply by the constant a , and both velocity and position can be found by using the following formulas:

a(t) = a  
v(t) = v 0 + at  
s(t) = s 0 + v 0 t + at 2  

Where v 0 is the initial velocity at time t = 0 and s 0 is the initial position at time t = 0 . Note that these formulas are in compliance with the relations v(t) = s'(t) and a(t) = v'(t) .

A ball dropped vertically from a height travels in this fashion, because it is accelerated by gravity at a constant rate of 9.8 meters per second per second.

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