For An Object Whose Velocity In $ft/sec$ Is Given By $v(t) = -t^2 + 2$, What Is Its Displacement, In Feet, On The Interval $ T = 0 T = 0 T = 0 [/tex] To $t = 2$ Seconds?1) 1.333 2) 2.438 3) 2.667 4) -4

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Understanding Displacement and Velocity

Displacement and velocity are two fundamental concepts in physics that describe the motion of an object. While velocity is a measure of the rate of change of an object's position with respect to time, displacement is the actual distance traveled by the object from its initial to its final position. In this article, we will explore how to calculate the displacement of an object given its velocity function.

The Velocity Function

The velocity function of an object is given by $v(t) = -t^2 + 2$, where $t$ is time in seconds and $v(t)$ is the velocity in feet per second. This function represents the rate of change of the object's position with respect to time.

Calculating Displacement

To calculate the displacement of the object, we need to integrate the velocity function with respect to time. The displacement function $s(t)$ is given by the integral of the velocity function:

s(t)=∫v(t)dts(t) = \int v(t) dt

Substituting the velocity function $v(t) = -t^2 + 2$ into the integral, we get:

s(t)=∫(−t2+2)dts(t) = \int (-t^2 + 2) dt

Evaluating the integral, we get:

s(t)=−t33+2t+Cs(t) = -\frac{t^3}{3} + 2t + C

where $C$ is the constant of integration.

Finding the Displacement on the Interval [0, 2]

To find the displacement of the object on the interval $[0, 2]$, we need to evaluate the displacement function $s(t)$ at $t = 2$ and subtract the value of $s(t)$ at $t = 0$.

First, we evaluate $s(t)$ at $t = 2$:

s(2)=−(2)33+2(2)+Cs(2) = -\frac{(2)^3}{3} + 2(2) + C

s(2)=−83+4+Cs(2) = -\frac{8}{3} + 4 + C

s(2)=−83+123+Cs(2) = -\frac{8}{3} + \frac{12}{3} + C

s(2)=43+Cs(2) = \frac{4}{3} + C

Next, we evaluate $s(t)$ at $t = 0$:

s(0)=−(0)33+2(0)+Cs(0) = -\frac{(0)^3}{3} + 2(0) + C

s(0)=0+Cs(0) = 0 + C

s(0)=Cs(0) = C

Now, we subtract the value of $s(t)$ at $t = 0$ from the value of $s(t)$ at $t = 2$:

s(2)−s(0)=43+C−Cs(2) - s(0) = \frac{4}{3} + C - C

s(2)−s(0)=43s(2) - s(0) = \frac{4}{3}

Therefore, the displacement of the object on the interval $[0, 2]$ is $\frac{4}{3}$ feet.

Conclusion

In this article, we have shown how to calculate the displacement of an object given its velocity function. We have used the integral of the velocity function to find the displacement function and then evaluated the displacement function at the endpoints of the interval to find the displacement of the object. The displacement of the object on the interval $[0, 2]$ is $\frac{4}{3}$ feet.

Answer

The correct answer is:

  1. 1.333 is incorrect
  2. 2.438 is incorrect
  3. 2.667 is incorrect
  4. -4 is incorrect

Q: What is the difference between velocity and displacement?

A: Velocity is a measure of the rate of change of an object's position with respect to time, while displacement is the actual distance traveled by the object from its initial to its final position.

Q: How do I calculate the displacement of an object given its velocity function?

A: To calculate the displacement of an object given its velocity function, you need to integrate the velocity function with respect to time. The displacement function is given by the integral of the velocity function.

Q: What is the formula for calculating displacement?

A: The formula for calculating displacement is:

s(t)=∫v(t)dts(t) = \int v(t) dt

where $s(t)$ is the displacement function and $v(t)$ is the velocity function.

Q: How do I evaluate the displacement function at the endpoints of the interval?

A: To evaluate the displacement function at the endpoints of the interval, you need to substitute the values of $t$ into the displacement function and simplify.

Q: What is the constant of integration in the displacement function?

A: The constant of integration in the displacement function is denoted by $C$ and is a constant that is added to the displacement function to make it a specific function.

Q: How do I find the displacement of an object on a given interval?

A: To find the displacement of an object on a given interval, you need to evaluate the displacement function at the endpoints of the interval and subtract the value of the displacement function at the initial point from the value of the displacement function at the final point.

Q: What is the significance of the displacement of an object?

A: The displacement of an object is a measure of the distance traveled by the object from its initial to its final position. It is an important concept in physics and is used to describe the motion of objects.

Q: Can I use the displacement function to find the velocity function?

A: No, you cannot use the displacement function to find the velocity function. The velocity function is the derivative of the displacement function, and you need to use the velocity function to find the displacement function.

Q: What is the relationship between displacement and velocity?

A: The displacement of an object is related to its velocity by the equation:

s(t)=∫v(t)dts(t) = \int v(t) dt

This equation shows that the displacement of an object is the integral of its velocity function.

Q: Can I use the displacement function to find the position function?

A: Yes, you can use the displacement function to find the position function. The position function is the integral of the displacement function, and you can use the displacement function to find the position function.

Q: What is the significance of the position function?

A: The position function is a measure of the position of an object at a given time. It is an important concept in physics and is used to describe the motion of objects.

Q: Can I use the displacement function to find the acceleration function?

A: No, you cannot use the displacement function to find the acceleration function. The acceleration function is the derivative of the velocity function, and you need to use the velocity function to find the acceleration function.

Q: What is the relationship between displacement, velocity, and acceleration?

A: The displacement, velocity, and acceleration of an object are related by the following equations:

s(t)=∫v(t)dts(t) = \int v(t) dt

v(t)=dsdtv(t) = \frac{ds}{dt}

a(t)=dvdta(t) = \frac{dv}{dt}

These equations show that the displacement of an object is the integral of its velocity function, the velocity of an object is the derivative of its displacement function, and the acceleration of an object is the derivative of its velocity function.