### Velocity#### Problem 1The position of an object is given as a function of time as \( x(t) = (3.0 \, \text{m/s})t + (2.0 \, \text{m/s}^2)t^2 \).**What is the average velocity of the object between \( t = 0.0 \, s \) and \( t = 2.0 \, s \)?**---**Solution:**1. First, find the position of the object at \( t = 0.0 \, s \): \[ x(0) = (3.0 \, \text{m/s})(0) + (2.0 \, \text{m/s}^2)(0)^2 = 0 \, \text{m} \]2. Next, find the position of the object at \( t = 2.0 \, s \): \[ x(2) = (3.0 \, \text{m/s})(2) + (2.0 \, \text{m/s}^2)(2)^2 \] \[ x(2) = (3.0 \, \text{m/s})(2) + (2.0 \, \text{m/s}^2)(4) \] \[ x(2) = 6.0 \, \text{m} + 8.0 \, \text{m} = 14.0 \, \text{m} \]3. The average velocity (\( \overline{v} \)) between \( t = 0.0 \, s \) and \( t = 2.0 \, s \) is given by the change in position divided by the change in time: \[ \overline{v} = \frac{x(2) - x(0)}{2.0 \, \text{s} - 0.0 \, \text{s}} \] \[ \overline{v} = \frac{14.0 \, \text{m} - 0.0 \, \text{m}}{2.0 \, \text{s}} \] \[ \overline{v} = \

Given that:Position of the object Time range

Answered: Velocity Problem 1. The position of an…

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