Calculating the Distance and Velocity of a Stone Thrown Toward a Toy

Introduction

In this article, we will explore the mathematical calculations involved in determining the distance of a stone from a toy over time and its resultant velocity. We will employ basic principles of calculus and algebra to solve the given problem. This is highly useful for understanding motion and its parameters in physics and can be of great help for students and professionals dealing with optimization, navigation, and mechanical systems.

Problem Description

A stone is thrown toward a toy from a distance of 120 feet. The position of the stone at any time t can be represented by the equation:

The distance from the stone to the toy at time t is given by: [ h(t) 2t^2 - 32t 120. ]

Understanding the Equation

The equation h(t) 2t2 - 32t 120 is a quadratic function, indicating the height or distance of the stone from the toy over time. Here, the coefficient of t2 is positive (2), which means the parabola opens upwards, indicating that the stone will initially fall towards the toy and eventually go past it.

Determining the Distance at Specific Times

At t 2 seconds:

[ h(2) 2(2)^2 - 32(2) 120 8 - 64 120 64 ] feet.

At t 5 seconds:

[ h(5) 2(5)^2 - 32(5) 120 50 - 160 120 10 ] feet.

Calculating the Instantaneous and Average Velocity

The velocity of the stone can be determined by taking the first derivative of the distance function with respect to time:

[ v(t) frac{dh}{dt} 4t - 32 ] feet per second.

At t 2 seconds:

[ v(2) 4(2) - 32 8 - 32 -24 ] feet per second.

At t 5 seconds:

[ v(5) 4(5) - 32 20 - 32 -12 ] feet per second.

For average velocity over the interval from t 2 seconds to t 5 seconds:

[ text{Average velocity} frac{h(5) - h(2)}{5 - 2} frac{10 - 64}{3} -18 ] feet per second.

The negative sign indicates that the stone is moving in the negative (decreasing) distance direction.

Ensuring the Stone Hits the Toy

Let's use the equation for distance to determine when the stone hits the toy:

[ 0 2t^2 - 32t 120 ] Solving this quadratic equation:

[ t frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where a 2, b -32, and c 120. [ t frac{32 pm sqrt{1024 - 960}}{4} ] [ t frac{32 pm sqrt{64}}{4} ] [ t frac{32 pm 8}{4} ] [ t 10 ] or [ t 6 ] Since t 10 would mean the stone has already passed the toy, we consider t 6 seconds as the point when the stone hits the toy.

Final Calculation

At t 6 seconds, the velocity is given by:

[ v(6) 4(6) - 32 24 - 32 -8 ] feet per second.

With the stone hitting the toy at t 6 seconds, we can now calculate the average velocity from t 2 seconds to t 6 seconds:

[ text{Average velocity} frac{h(6) - h(2)}{6 - 2} frac{0 - 64}{4} -16 ] feet per second.

Thus, the stone does indeed hit the toy, and the average velocity in this interval is -16 feet per second.

Conclusion

This comprehensive analysis not only explains the fundamental concepts of distance and velocity in physics but also demonstrates how calculus can be utilized in real-world scenarios, such as determining the motion of objects in space. By understanding the principles outlined in this article, one can effectively apply these concepts to a wide range of problems involving motion and its parameters.