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). You are given: x(7)=6quad" and "quad v(7)=3 Which of the following expression gives the velocity of the particle when t=8? int_(8)^(7)a(t)dt 3+int_(8)^(7)a(t)dt int_(7)^(8)a(t)dt 3+int_(7)^(8)a(t)dt

Given Data: We are given the position and velocity of the particle at t=7, which are x(7)=6 and v(7)=3, respectively. To find the velocity at t=8, we need to consider the change in velocity from t=7 to t=8, which is given by the integral of the acceleration function a(t) over the interval from t=7 to t=8.Change in Velocity: The fundamental theorem of calculus tells us that the integral of the acceleration function a(t) from time t=7 to t=8 will give us the change in velocity over that time interval. Since we are given the velocity at t=7 as v(7)=3, we need to add this initial velocity to the change in velocity to find the velocity at t=8.Fundamental Theorem of Calculus: The correct expression for the velocity at t=8 is the initial velocity at t=7 plus the integral of the acceleration function from t=7 to t=8. Mathematically, this is represented as v(8) = v(7) + ∫(from t=7 to t=8) a(t) dt.Expression for Velocity: Therefore, the correct expression that gives the velocity of the particle when t=8 is 3 + ∫(from t=7 to t=8) a(t) dt. This corresponds to the expression ＂3+int_(7)^(8)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). You are given:

x(7)=6quad" and "quad v(7)=3
Which of the following expression gives the velocity of the particle when
t=8?

int_(8)^(7)a(t)dt

3+int_(8)^(7)a(t)dt

int_(7)^(8)a(t)dt

3+int_(7)^(8)a(t)dt

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 $a(t)$ . You are given: $\newline$ $x(7)=6 \quad \text { and } \quad v(7)=3$ $\newline$ Which of the following expression gives the velocity of the particle when $t=8 ?$ $\newline$ $\int_{8}^{7} a(t) d t$ $\newline$ $3+\int_{8}^{7} a(t) d t$ $\newline$ $\int_{7}^{8} a(t) d t$ $\newline$ $3+\int_{7}^{8} a(t) d t$

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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)

. You are given:

\newline

x(7)=6 \quad \text { and } \quad v(7)=3

\newline

Which of the following expression gives the velocity of the particle when

t=8 ?

\newline

\int_{8}^{7} a(t) d t

\newline

3+\int_{8}^{7} a(t) d t

\newline

\int_{7}^{8} a(t) d t

\newline

3+\int_{7}^{8} a(t) d t

Given Data: We are given the position and velocity of the particle at $t=7$ , which are $x(7)=6$ and $v(7)=3$ , respectively. To find the velocity at $t=8$ , we need to consider the change in velocity from $t=7$ to $t=8$ , which is given by the integral of the acceleration function $a(t)$ over the interval from $t=7$ to $t=8$ .
Change in Velocity: The fundamental theorem of calculus tells us that the integral of the acceleration function $a(t)$ from time $t=7$ to $t=8$ will give us the change in velocity over that time interval. Since we are given the velocity at $t=7$ as $v(7)=3$ , we need to add this initial velocity to the change in velocity to find the velocity at $t=8$ .
Fundamental Theorem of Calculus: The correct expression for the velocity at $t=8$ is the initial velocity at $t=7$ plus the integral of the acceleration function from $t=7$ to $t=8$ . Mathematically, this is represented as $v(8) = v(7) + \int_{t=7}^{t=8} a(t) \, dt$ .
Expression for Velocity: Therefore, the correct expression that gives the velocity of the particle when $t=8$ is $3 + \int_{t=7}^{t=8} a(t) \, dt$ . This corresponds to the expression $3+\int_{7}^{8}a(t)\,dt$ .