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Since up is positive, the final position of the rock will be negative because it finishes below the starting point at \(y_0=0\). Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward. Solution. 1.
Solution. When you project an object upward and release it at its initial velocity, it is moving in the opposite direction of the force of gravity. Thus the initial velocity is negative. The velocity of the object is also negative on the way up but positive on the way down. Note: The convention we use is that upward velocities are negative and ...
For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of[latex]\boldsymbol{\pm3.20\textbf{ m/s}}[/latex]is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when ...
- OpenStax
- 2016
Jun 15, 2015 · Negative velocity just means velocity in the opposite direction than what would be positive. At the core of it, signs have no meaning in real life. Directions have meaning, and signs are a mathematical way to indicate or alter 1-dimensional direction. Similarly, time can be positive (tomorrow) and negative (yesterday).
Velocity can be negative in physics or calculus, such as when an object is in free fall (due to the effects of gravity). Speed is the absolute value of velocity, which gives us a nonnegative number with no direction. Velocity is negative whenever position is decreasing. This is because velocity is the derivative of position.
For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down.
For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 2.14) when the initial velocity is 13.0 m/s straight up, a result of ± 3. 20 m/s ± 3. 20 m/s is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 ...
