An Object Moves Along A Straight Line So That At Any Time T T T , Its Acceleration Is Given By Α ( T ) = 15 T \alpha(t) = 15 \sqrt{t} Α ( T ) = 15 T ​ . At Time T = 0 T = 0 T = 0 , The Object's Position Is 10, And At Time T = 1 T = 1 T = 1 , The Object's Position Is 20.

Introduction

In physics, the motion of an object is often described using mathematical equations that relate its position, velocity, and acceleration over time. In this article, we will explore the motion of an object that moves along a straight line, with its acceleration given by a specific function of time. We will use this information to determine the object's position at any given time, and discuss the implications of this motion.

Acceleration and Velocity

The acceleration of an object is defined as the rate of change of its velocity. Mathematically, this can be expressed as:

$$\alpha(t) = \frac{dv}{dt}$$

where $\alpha(t)$ is the acceleration at time $t$, and $v$ is the velocity of the object.

In this case, the acceleration is given by the function:

$$\alpha(t) = 15 \sqrt{t}$$

This means that the acceleration of the object is proportional to the square root of time. To find the velocity of the object, we need to integrate the acceleration function with respect to time:

$$v(t) = \int \alpha(t) dt = \int 15 \sqrt{t} dt$$

Evaluating this integral, we get:

$$v(t) = 15 \left( \frac{2}{3} t^{\frac{3}{2}} \right) + C$$

where $C$ is the constant of integration. Since the initial velocity is not given, we can set $C = 0$.

Position and Velocity

The position of an object is defined as the distance it has traveled from a reference point. Mathematically, this can be expressed as:

$$s(t) = \int v(t) dt$$

where $s(t)$ is the position at time $t$, and $v(t)$ is the velocity of the object.

Substituting the expression for velocity, we get:

$$s(t) = \int 15 \left( \frac{2}{3} t^{\frac{3}{2}} \right) dt$$

Evaluating this integral, we get:

$$s(t) = 10 t^{\frac{5}{2}} + D$$

where $D$ is the constant of integration. Since the initial position is given as 10, we can set $D = 10$.

Solving for the Position Function

We are given that at time $t = 0$, the object's position is 10, and at time $t = 1$, the object's position is 20. We can use this information to solve for the constant of integration, $D$.

Substituting $t = 0$ into the position function, we get:

$$s(0) = 10 (0)^{\frac{5}{2}} + D = 10$$

This confirms that the initial position is indeed 10.

Substituting $t = 1$ into the position function, we get:

$$s(1) = 10 (1)^{\frac{5}{2}} + D = 20$$

Solving for $D$, we get:

$$D = 20 - 10 = 10$$

The Final Position Function

Now that we have solved for the constant of integration, $D$, we can write the final position function:

$$s(t) = 10 t^{\frac{5}{2}} + 10$$

This function describes the position of the object at any given time, $t$.

Conclusion

In this article, we have explored the motion of an object that moves along a straight line, with its acceleration given by a specific function of time. We have used this information to determine the object's position at any given time, and discussed the implications of this motion. The final position function, $s(t) = 10 t^{\frac{5}{2}} + 10$, describes the position of the object at any given time, $t$.

Future Work

In future work, we can explore the motion of objects with more complex acceleration functions, and discuss the implications of these motions on the object's position and velocity.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Additional Resources

• [1] Khan Academy. (n.d.). Motion in One Dimension. Retrieved from https://www.khanacademy.org/science/physics/motion-in-one-dimension
• [2] MIT OpenCourseWare. (n.d.). Physics 8.01: Classical Mechanics. Retrieved from https://ocw.mit.edu/courses/physics/8-01-classical-mechanics-fall-2005/
  

Introduction

In our previous article, we explored the motion of an object that moves along a straight line, with its acceleration given by a specific function of time. We determined the object's position at any given time, and discussed the implications of this motion. In this article, we will answer some common questions related to this topic.

Q: What is the acceleration of the object at time t = 0?

A: The acceleration of the object at time t = 0 is given by the function:

$$\alpha(t) = 15 \sqrt{t}$$

Substituting t = 0 into this function, we get:

$$\alpha(0) = 15 \sqrt{0} = 0$$

So, the acceleration of the object at time t = 0 is 0.

Q: What is the velocity of the object at time t = 0?

A: The velocity of the object at time t = 0 is given by the function:

$$v(t) = 15 \left( \frac{2}{3} t^{\frac{3}{2}} \right)$$

Substituting t = 0 into this function, we get:

$$v(0) = 15 \left( \frac{2}{3} (0)^{\frac{3}{2}} \right) = 0$$

So, the velocity of the object at time t = 0 is 0.

Q: What is the position of the object at time t = 0?

A: The position of the object at time t = 0 is given as 10.

Q: What is the acceleration of the object at time t = 1?

A: The acceleration of the object at time t = 1 is given by the function:

$$\alpha(t) = 15 \sqrt{t}$$

Substituting t = 1 into this function, we get:

$$\alpha(1) = 15 \sqrt{1} = 15$$

So, the acceleration of the object at time t = 1 is 15.

Q: What is the velocity of the object at time t = 1?

A: The velocity of the object at time t = 1 is given by the function:

$$v(t) = 15 \left( \frac{2}{3} t^{\frac{3}{2}} \right)$$

Substituting t = 1 into this function, we get:

$$v(1) = 15 \left( \frac{2}{3} (1)^{\frac{3}{2}} \right) = 10$$

So, the velocity of the object at time t = 1 is 10.

Q: What is the position of the object at time t = 1?

A: The position of the object at time t = 1 is given as 20.

Q: How can we use the position function to determine the object's position at any given time?

A: We can use the position function:

$$s(t) = 10 t^{\frac{5}{2}} + 10$$

to determine the object's position at any given time, t.

Q: What is the significance of the constant of integration, D?

A: The constant of integration, D, represents the initial position of the object. In this case, D = 10.

Q: How can we use the acceleration function to determine the object's acceleration at any given time?

A: We can use the acceleration function:

$$\alpha(t) = 15 \sqrt{t}$$

to determine the object's acceleration at any given time, t.

Conclusion

In this article, we have answered some common questions related to the motion of an object that moves along a straight line, with its acceleration given by a specific function of time. We have discussed the implications of this motion and provided a position function that describes the object's position at any given time.

Future Work

In future work, we can explore the motion of objects with more complex acceleration functions, and discuss the implications of these motions on the object's position and velocity.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics (10th ed.). John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage Learning.

Additional Resources

• [1] Khan Academy. (n.d.). Motion in One Dimension. Retrieved from https://www.khanacademy.org/science/physics/motion-in-one-dimension
• [2] MIT OpenCourseWare. (n.d.). Physics 8.01: Classical Mechanics. Retrieved from https://ocw.mit.edu/courses/physics/8-01-classical-mechanics-fall-2005/