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Rock Hoffman Minh Pham
Megan Seymour “The Car and the Truck” Write Up
The problem called “The Car and the Truck” deals with a car and truck travelling along a
straight road at different times and different velocities. The car starts an hour earlier than the
truck and has a velocity that is indicated by the velocity (mph) vs. time (hour) graph on page
317, which starts out concave down and increasing until t= 2 hours, when the derivative is equal
to zero, and the rest of the graph is decreasing. An hour later, at t=1,the truck starts to travel at
a constant velocity of 50 mph. The problem first asks the distance that the car has travelled in
the first hour. Then it asks the rate of change, or the derivative, of the distance between the car
and the truck at t = 3 hours, in addition to the real life significance of the fact that the car’s
velocity is maximized at t = 2 hours. Next, the question asks to find when the distance between
the car and the truck is the greatest and what that distance is. The problem the goes on to ask
when the truck overtakes the car (in time) and at what distance from the starting point. Finally,
the problem hypothetically proposes that both vehicles start at the same time and asks for the
graph of the two functions when both having initial values of (0,0) and to determine how many
times the truck and car pass each other and what it means in terms of the distance between the
two pass one another. One must use his or her knowledge of area under the curve of velocity
graphs, the Fundamental Theorem of Calculus, integrals, and critical thinking to answer this
problem accurately.
Concepts:
The way to calculate distance from a velocity equation is by taking the integral of that
equation. The velocity equation is the derivative of the position function. The integral is
essentially the area under the graph. To calculate this algebraically, one must take the
antiderivative of the equation, but in the case of this problem, since the graph is given, one must
solve the problem graphically.
The graphical was to correctly execute taking an Integral to get the total distance
traveled is by looking at the area under the velocity curve. One can estimate this through taking
Riemann's sums or just counting the number of boxes under the curve. A Riemann’s sum can
be an overestimate or an underestimate depending on the nature of the graph and what kind of
sum the student chooses to use.
To take a Riemann sum:
The student must first split the graph into intervals, which don’t have to necessarily be
equal, but they have to be known. These intervals will be known as t n , and knowing them is
essential in taking Riemann sums. Then, for left hand sums one takes the f(t) value from the left
side of the interval and multiplies it by the interval. One has to execute this for all of the intervals
and then the sum will be complete. For a right hand sum one would do the same thing but using
the f(t) value from the right side of the interval.
A general Riemann sum for f on the interval [a,b] is a sum of the form:
f(c i ) i ,∑n
i=1tΔ
Where 1 is the initial (starting) point and n is the end point. is the value of interval, i, and f(c i )tΔ
is the output value at this interval, which can be any point on this interval, whether leftmost,
rightmost, or even mid point.
This is how one sets up an integral:
Suppose f is continuous for . The definite integral of f from a to b can be written a ≤ t ≤ b
as, (t)dt∫b
af
However, one must remember, in order to get the correct distance that is being asked
for, one must determine the bounds of this Integral. Or, graphically speaking, one must know the
beginning (left endpoint) and the end (right endpoint) of the portion of the velocity graph their are
seeking to integrate.
Solutions:
a. When the truck starts, the car has traveled 1 hour already. The distance traveled by the car can be found by the total area in the velocity graph from 0 to 1 hour, which is 7 rectangles. Each triangle is 5 miles; therefore, the total distance is 35 hours.
b. Velocity of car at 3pm: 67 mph
Velocity of truck at 3pm: 50 mph We call function f(x) as the distance between the car and the truck: F(x) = f(car) – f(truck) The rate of change of the distance between the car and the truck will be the derivative of the function f(x), which is the difference in velocity. F’(x) = f’(Car) – f’(truck) F’(x) = v(car) – v(truck) F’(3) = 67 50 = 17 mph Therefore, at 3pm, the distance between the car and the truck is increasing at 17 miles per hour.
When the car’s velocity is maximized at 2pm, the rate of the distance between the car and the truck (F’(x) = f’(Car) – f’(truck)) will be the greatest since the truck’s velocity stays constant and so the difference between them will be the greatest.
c. When the car is ahead of the truck, the distance between the car and the truck will keep increasing until the truck’s velocity catches up with the car’s velocity. Therefore, the greatest distance will occur when v (truck)= v(car). Since v (truck) is always 50mph, v (car) will be 50mph at t = 4.3 hours (4:15pm). The distance traveled by the car from t=0 to t=4.3 is the area under the v (car) graph, which is around 50 squares; therefore the distance is 250 miles
The distance traveled by the truck from t= 1 to t= 4.3 is 50 * 3.3 = 165 miles (33 squares)
Therefore, the greatest distance is 250 165 = 85 miles (17 squares)
d. The truck overtakes the car when the distance traveled by the car and the truck are
equal. This means that the area under the curve v(car) and the area under the curve v(truck) are equal. Until the time 4.3, the distance of the car and the truck are the greatest apart. After t =4.3, the truck starts catching up to the car, and so the gap between their distance get smaller. Under the graph below, the red shaded part represents the total distance apart between the car and the truck until t = 4.3. To cancel
it out, we have to find the area that is equal to that total distance after t = 4.3 so that the differences can cancel out. The corresponding time will be the time where the car and the truck have traveled the same distance. The number of squares before t= 4.3: 17 squares Therefore, the time will be: t = 8.3 (where the 17 squares after 4.3) = 8:25pm
e.
f. The graphs intersect twice, at t = 0.7 and 4.3 hours. The first intersection point (t = 0.7) means that v(car) = v(truck). Before t=0.7,
v(truck) is greater than v(car), which means that the truck is traveling away from the
car car. After t= 0.7, v(car) becomes greater than v(truck), which means that the car is catching up with the truck and closing the distance between them.
The second intersection point (t = 4.3) also means that v(car) = v(truck). At this point, the distance apart between the car and the truck are the greatest. Before t =4.3, v(car) is greater than v(truck), which means that the car is traveling away from the truck. After t= 0.7, v(truck) becomes greater than v(car), which means that the truck is catching up with the car and closing the distance between them.
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