I NEED A RESPONSE FOR THIS DISCUSSION ( 30 - 50 WORDS ) Average velocity can be
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Question
I NEED A RESPONSE FOR THIS DISCUSSION ( 30 - 50 WORDS )
Average velocity can be defined as a distance traveled by an object divided by the amount of time the object traveled. For example, if a car travels 20 meters in 10 seconds, we can quickly calculate the average velocity using our formula x/t. Using the starting point (0 m) at time 0 and the given information, we know that we can calculate x, by subtracting the first distance from the second (20 m - 0 m). Then, we will divide that answer by t, which can be calculated by subtracting the first time value from the second (10 s - 0 s). This will give us the average velocity, 2 m/s.
Instantaneous Velocity can be defined as the velocity of an object at a specific point of time. We can find this by finding the limit as t approaches zero of d/dt. For example, if the position can be found by the equation 4x^2 + 3x + 1, we would need to take the derivative of this equation to find the velocity, which is 8x+3. Then, we could plug in any x-value to find the instantaneous velocity at that point. If we wanted to find the instantaneous velocity at x=1, then our final answer would be 11 m/s for this equation.
Average acceleration can be defined as the rate that velocity changes divided by a change of time. Using the equation v/t, we can find the average acceleration. For example, if we know that our car's starting velocity was 10 m/s at 5 seconds and our second velocity was 20 m/s at 10 seconds, we could plug in these values to find the average acceleration ((20 m/s-10 m/s)/(10 s- 5 s)). This gives us an average acceleration of 2 m/s^2.
Instantaneous acceleration can be defined as the rate that the velocity divided by time changes at a certain instant. We can find this by taking the limit as t approaches zero of d^2v/d^2t. For example, if a car's position function is 2x^3 + 3x^2 + 10x, we need to take the second derivative of the function to get to the acceleration, which is 12x+6. Then, we would plug in the certain x-value we are looking for. If the x-value we want is 1, then our instantaneous acceleration at that point would be 18 m/s^2.
To review the problems in the textbook, I found that as long as we kept these formulas in mind, the questions were easy to answer, as we would only need to plug in our missing values. In my own experience, I found that it was very important for me to keep track of labels when doing these calculations. This helped me to keep track of my units!
Explanation / Answer
In my opinion your understanding of average and instantaneous velocity & acceleration is pretty good.
Just wanted to add to instantaneos velocity and acceleration. We get it from derivatives. A derivative, e.g. dy/dx gives rate of change of y wrt x. Here y is a function of x.
Similarly instantaneous velocity is calculated by differentiating position wrt time. here position is a function of time. In your example the equation of position is given as 4x^2 + 3x + 1 and instantaneous velocity is asked at x = 1. Instead this equation should be 4t^2 + 3t + 1 and instantaneous velocity should be asked at t = 1. Instatntaneous velocity can be found by calculating d/dt(4t^2 + 3t + 1) which will come out to be 8t + 3. And after putting t = 1, we can get the value of velocity = 11 m/s at that particular time t = 1.
Same explanation applies for instantaneous acceleration.
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