How do you find a speed when you are given only the acceleration?

Integrate the acceleration to get the velocity, using a known velocity to fix the constant, then take the absolute value. A question asking about speed rather than velocity always needs that last step.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 8: Applications of IntegrationDefinite Integrals in Context

Definite Integrals in Context (College Board AP® Calculus BC): Flashcards

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FrontAverage Value of a Function
Fill in the two missing parts of the average value of $f$ over $[a, b]$ :
$\frac{1}{______} \int_{a}^{b} f (x)______$ for a continuous function $f$

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Cards in this collection (28)

Fill in the two missing parts of the average value of $f$ over $[a, b]$ :
$\frac{1}{______} \int_{a}^{b} f (x)______$ for a continuous function $f$
The completed formula is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ .
The integral gives the accumulated change, and dividing by the width of the interval turns that into an average.
How does average value differ from average rate of change?
Average value is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ , an integral.
Average rate of change is $\frac{f (b) - f (a)}{b - a}$ , a difference quotient, and the two are quite different quantities.
True or False?
A continuous function actually attains its own average value somewhere on the interval.
True.
This is the mean value theorem for integrals: there is some $c$ in $[a, b]$ with $f (c)$ equal to the average value.
So the average value is a value the function really takes, not merely a number computed from it.
What does the average value mean geometrically?
The constant function $g (x) = k$ encloses the same area over $[a, b]$ as $f$ does.
A rectangle of height $k$ and width $b - a$ has area $k (b - a)$ , which equals $\int_{a}^{b} f (x) d x$ .
Find the average value of $f (x) = \sin x$ over $[0, π]$ .
It is $\frac{2}{π}$ .
The formula gives $\frac{1}{π} {[- \cos x]}_{0}^{π} = \frac{1}{π} (1 + 1)$ .
Can you speak of the average value of a function in general?
No.
The average value is only defined for a particular interval $[a, b]$ , and it usually changes when the interval does.
A rate of flow in gallons per minute has average value 12 over $[0, 5]$ . What does that mean?
The tank filled at an average rate of 12 gallons per minute across those five minutes.
The same total volume would have arrived from a constant flow of 12 gallons per minute for the whole interval.
Fill in the missing word about what a definite integral of a rate gives:
the definite integral of the rate of change of a quantity over an interval gives the $______$ change of that quantity over that interval
The completed statement is: the definite integral of the rate of change of a quantity over an interval gives the net change of that quantity over that interval.
It sums the infinitesimal changes $f (x) d x$ right across the interval.
Define marginal cost.
Marginal cost is $C^{'} (x)$ , the rate at which cost changes as one more unit is sold.
Marginal revenue and marginal profit are $R^{'} (x)$ and $P^{'} (x)$ in the same way.
A company's marginal cost is $C^{'} (x) = 300 + 0.02 x$ dollars per set. Interpret $C^{'} (200) = 304$ .
When 200 sets have been sold, cost is increasing at 304 dollars per additional set.
A positive rate of change means the cost is rising, and the units are those of the quantity divided by those of the variable.
True or False?
If food is eaten from a bowl at rate $f (t)$ , the amount left is $F (0) + \int_{0}^{t} f (s) d s$ .
False.
The rate of eating is the rate at which food leaves the bowl, so the amount remaining is $F (0) - \int_{0}^{t} f (s) d s$ .
Always check which quantity the given rate is the rate of change of.
With $C^{'} (x) = 300 + 0.02 x$ , find the change in cost from $x = 200$ to $x = 300$ .
The cost rises by 30 500 dollars.
Evaluating $\int_{200}^{300} (300 + 0.02 x) d x = {[300 x + 0.01 x^{2}]}_{200}^{300}$ gives $90900 - 60400$ .
If $P = R - C$ , how do you find the change in profit from the marginals?
Integrate $P^{'} (x) = R^{'} (x) - C^{'} (x)$ over the interval.
Differentiating a difference gives the difference of the derivatives, so the marginals subtract just as the quantities themselves do.
Food is eaten at a rate $f (t) = 4 + 4 \cos (\frac{π}{4} t)$ grams per hour. How much is eaten in the first four hours?
16 grams.
Evaluating $\int_{0}^{4} (4 + 4 \cos (\frac{π}{4} t)) d t = 4 {[t + \frac{4}{π} \sin (\frac{π}{4} t)]}_{0}^{4}$ gives $4 (4 + 0)$ .
Fill in the two missing quantities in the integral relations for motion:
$v (t) = \int______d t$ and $s (t) = \int______d t$ for a particle on a line
The completed relations are $v (t) = \int a (t) d t$ and $s (t) = \int v (t) d t$ .
Each reverses a derivative: acceleration is the rate of change of velocity, and velocity the rate of change of displacement.
What does $\int_{t_{1}}^{t_{2}} a (t) d t$ represent?
The total change in velocity between those two times.
It is the area under the acceleration-time graph, and it is not the final velocity.
True or False?
$\int_{t_{1}}^{t_{2}} v (t) d t$ gives the particle's position at time $t_{2}$ .
False.
It gives the change in displacement over the interval, not the final position.
To reach the position at $t_{2}$ you add that change to the known displacement at $t_{1}$ .
A particle has $a (t) = \frac{1}{4} \sqrt{t}$ and velocity 2 at $t = 3$ . Find its velocity at $t = 45$ .
About $51.446$ .
The change in velocity is $\int_{3}^{45} \frac{1}{4} \sqrt{t} d t = \frac{1}{6} (45^{\frac{3}{2}} - 3^{\frac{3}{2}}) = 49.446$ , and adding the starting value 2 gives the answer.
What are the two ways to get an expression for displacement from velocity?
Either write $s (t) = s (t_{0}) + \int_{t_{0}}^{t} v (w) d w$ with a dummy variable, or find the indefinite integral and fix the constant from a known displacement.
Both give the same expression.
A particle has $v (t) = 12 t - \frac{1}{20} t^{3}$ and displacement 40 at $t = 3$ . Find its displacement at $t = 10$ .
About $462.013$ .
The change is $\int_{3}^{10} (12 t - \frac{1}{20} t^{3}) d t = {[6 t^{2} - \frac{1}{80} t^{4}]}_{3}^{10} = 422.0125$ , and adding 40 gives the answer.
A particle starts at displacement zero. How do you find when it next returns there?
Solve $\int_{0}^{T} v (t) d t = 0$ for the upper limit $T$ .
The root $T = 0$ is the start of the motion itself, so the answer is the other root.
Fill in the missing function in the total distance traveled between two times:
$\int_{t_{1}}^{t_{2}}______d t$ for a particle with velocity $v (t)$
The completed integral is $\int_{t_{1}}^{t_{2}} | v (t) | d t$ .
The absolute value is what makes a stretch of backwards motion add to the distance instead of subtracting from it.
Why do displacement and distance need different integrals?
Because a negative velocity decreases the displacement while still increasing the distance traveled.
So displacement uses $\int v (t) d t$ and distance uses $\int | v (t) | d t$ .
For $v = 2 t - 6$ on $[0, 6]$ , find the displacement and the distance.
The displacement is $0$ and the distance is $18$ .
The integral is $- 9$ from 0 to 3 and $+ 9$ from 3 to 6, so the two cancel for displacement but add to 18 for distance.
True or False?
A particle whose velocity increases by 10 must have gained speed.
False.
If the velocity goes from $- 8$ to $2$ it has increased by 10, yet the speed has fallen from 8 to 2.
A change of velocity that passes through zero can reduce the speed.
How do you evaluate $\int | v (t) | d t$ without a calculator?
Find where $v$ crosses zero and split the integral at those times.
Evaluate each piece separately, then make any negative result positive before adding them all together.
A particle has $v (t) = 3 \sin \frac{t}{2}$ . Find the total distance traveled between $t = π$ and $t = 4 π$ .
Total distance traveled = 18.
The velocity is zero at $2 π$ , so split there: the integral is $6$ on $[π, 2 π]$ and $- 12$ on $[2 π, 4 π]$ .
Taking the absolute value of the second gives $6 + 12$ .
How do you find a speed when you are given only the acceleration?
Integrate the acceleration to get the velocity, using a known velocity to fix the constant, then take the absolute value.
A question asking about speed rather than velocity always needs that last step.

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