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(0)=1" and "v(0)=3 Which of the following expression gives the velocity of the particle when t=8? 1+int_(8)^(0)v(t)dt 3+int_(8)^(0)a(t)dt 1+int_(0)^(8)v(t)dt 3+int_(0)^(8)a(t)dt

Consider Initial Velocity: To find the velocity of the particle at t=8, we need to consider the initial velocity and the change in velocity over time. The change in velocity is given by the integral of the acceleration function a(t) from the initial time to t=8.Calculate Change in Velocity: We are given the initial velocity v(0)=3. This is the velocity of the particle at t=0. To find the velocity at t=8, we need to add the change in velocity from t=0 to t=8 to the initial velocity.Integrate Acceleration Function: The change in velocity from t=0 to t=8 is given by the integral of the acceleration function a(t) over the interval [0, 8]. This is represented mathematically as ∫ from 0 to 8 of a(t) dt.Find Velocity at t=8: Therefore, the correct expression for the velocity of the particle at t=8 is the initial velocity plus the integral of the acceleration from time 0 to 8. This is represented by the expression 3 + ∫ from 0 to 8 of a(t) dt.Correct Expression for Velocity: The correct choice from the given options that represents this expression is ＂3+int_(0)^(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(0)=1" and "v(0)=3
Which of the following expression gives the velocity of the particle when
t=8?

1+int_(8)^(0)v(t)dt

3+int_(8)^(0)a(t)dt

1+int_(0)^(8)v(t)dt

3+int_(0)^(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: $\newline$ $x(0)=1 \text { and } v(0)=3$ $\newline$ Which of the following expression gives the velocity of the particle when $t=8 ?$ $\newline$ $1+\int_{8}^{0} v(t) d t$ $\newline$ $3+\int_{8}^{0} a(t) d t$ $\newline$ $1+\int_{0}^{8} v(t) d t$ $\newline$ $3+\int_{0}^{8} a(t) d t$

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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(0)=1 \text { and } v(0)=3

\newline

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

t=8 ?

\newline

1+\int_{8}^{0} v(t) d t

\newline

3+\int_{8}^{0} a(t) d t

\newline

1+\int_{0}^{8} v(t) d t

\newline

3+\int_{0}^{8} a(t) d t

Consider Initial Velocity: To find the velocity of the particle at $t=8$ , we need to consider the initial velocity and the change in velocity over time. The change in velocity is given by the integral of the acceleration function $a(t)$ from the initial time to $t=8$ .
Calculate Change in Velocity: We are given the initial velocity $v(0)=3$ . This is the velocity of the particle at $t=0$ . To find the velocity at $t=8$ , we need to add the change in velocity from $t=0$ to $t=8$ to the initial velocity.
Integrate Acceleration Function: The change in velocity from $t=0$ to $t=8$ is given by the integral of the acceleration function $a(t)$ over the interval $[0, 8]$ . This is represented mathematically as $\int_{0}^{8} a(t) \, dt$ .
Find Velocity at $t=8$ : Therefore, the correct expression for the velocity of the particle at $t=8$ is the initial velocity plus the integral of the acceleration from time $0$ to $8$ . This is represented by the expression $3 + \int_{0}^{8} a(t) \, dt$ .
Correct Expression for Velocity: The correct choice from the given options that represents this expression is ＂ $3+\int_{0}^{8}a(t)\,dt$ ＂.