A particle moves along the x-axis such that at any time t >= 0 its position is x(t), its velocity is v(t), and its acceleration is a(t). What is the average acceleration of the particle on the interval 0 <= t <= 9? (1)/(9)int_(0)^(9)v(t)dt (x(9)-x(0))/(9) (1)/(9)int_(0)^(9)a(t)dt (a(9)-a(0))/(9)

Formula Application: To find the average acceleration over a time interval, we use the formula for average value of a function over an interval [a, b], which is given by: (1)/(b-a) * int_(a)^(b)f(t)dt In this case, the function f(t) is the acceleration a(t), and the interval is from t=0 to t=9.Substitution and Simplification: We substitute the given values into the formula for average acceleration: Average acceleration = (1)/(9-0) * int_(0)^(9)a(t)dt This simplifies to: Average acceleration = (1)/(9) * int_(0)^(9)a(t)dtElimination of Incorrect Options: The given options to represent the average acceleration are: (1)/(9)int_(0)^(9)v(t)dt (x(9)-x(0))/(9) (1)/(9)int_(0)^(9)a(t)dt (a(9)-a(0))/(9) We can immediately eliminate the first option because it represents the average velocity, not acceleration. The second option is incorrect because it represents the average velocity using position functions, not acceleration. The fourth option is incorrect because it represents the change in acceleration over time, not the average acceleration. The correct representation for the average acceleration is the third option: (1)/(9)int_(0)^(9)a(t)dt

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A particle moves along the
x-axis such that at any time
t >= 0 its position is
x(t), its velocity is
v(t), and its acceleration is
a(t).
What is the average acceleration of the particle on the interval
0 <= t <= 9?

(1)/(9)int_(0)^(9)v(t)dt

(x(9)-x(0))/(9)

(1)/(9)int_(0)^(9)a(t)dt

(a(9)-a(0))/(9)

A particle moves along the $x$ -axis such that at any time $t \geq 0$ its position is $x(t)$ , its velocity is $v(t)$ , and its acceleration is $a(t)$ . $\newline$ What is the average acceleration of the particle on the interval $0 \leq t \leq 9 ?$ $\newline$ $\frac{1}{9} \int_{0}^{9} v(t) d t$ $\newline$ $\frac{x(9)-x(0)}{9}$ $\newline$ $\frac{1}{9} \int_{0}^{9} a(t) d t$ $\newline$ $\frac{a(9)-a(0)}{9}$

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Math Problems

Calculus

Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. A particle moves along the

x

-axis such that at any time

t \geq 0

its position is

x(t)

, its velocity is

v(t)

, and its acceleration is

a(t)

\newline

What is the average acceleration of the particle on the interval

0 \leq t \leq 9 ?

\newline

\frac{1}{9} \int_{0}^{9} v(t) d t

\newline

\frac{x(9)-x(0)}{9}

\newline

\frac{1}{9} \int_{0}^{9} a(t) d t

\newline

\frac{a(9)-a(0)}{9}

Formula Application: To find the average acceleration over a time interval, we use the formula for average value of a function over an interval $[a, b]$ , which is given by: $\newline$ $\frac{1}{b-a} \int_{a}^{b}f(t)dt$ $\newline$ In this case, the function $f(t)$ is the acceleration $a(t)$ , and the interval is from $t=0$ to $t=9$ .
Substitution and Simplification: We substitute the given values into the formula for average acceleration: $\newline$ Average acceleration = $\frac{1}{9-0} \int_{0}^{9}a(t)dt$ $\newline$ This simplifies to: $\newline$ Average acceleration = $\frac{1}{9} \int_{0}^{9}a(t)dt$
Elimination of Incorrect Options: The given options to represent the average acceleration are: $\newline$ $\frac{1}{9}\int_{0}^{9}v(t)dt$ $\newline$ $\frac{x(9)-x(0)}{9}$ $\newline$ $\frac{1}{9}\int_{0}^{9}a(t)dt$ $\newline$ $\frac{a(9)-a(0)}{9}$ $\newline$ We can immediately eliminate the first option because it represents the average velocity, not acceleration. $\newline$ The second option is incorrect because it represents the average velocity using position functions, not acceleration. $\newline$ The fourth option is incorrect because it represents the change in acceleration over time, not the average acceleration. $\newline$ The correct representation for the average acceleration is the third option: $\newline$ $\frac{1}{9}\int_{0}^{9}a(t)dt$