Question

The height above the ground of a stone thrown upwards is given by​ s(t), where t is measured in seconds. After 1secondnothing​,the height of the stone is 47feet above the​ ground, and after 1.5​seconds, the height of the stone is 54feet above the ground. Evaluate ​s(1​)and ​s(1.5​),and then find the average velocity of the stone over the time interval ​[1​,1.5​].

Answer 1

Answer:

v = 14 ft/s

Explanation:

• By definition, the average velocity, is just the rate of change of the position, with respect to time, which can be written as follows:

       $v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x_{f}-x_{o}}{t_{f} - t_{o}} (1)$

• Defining the vertical position as the y-coordinate, with the origin at ground level, and the upward direction as positive, we can write (1) as follows:

       $v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} (2)$

• where yf = 54 ft, y₀ = 47 ft, tif = 1.5 s, t₀ = 1s.
• Replacing in (2) we get:

       $v_{avg} = \frac{\Delta y}{\Delta t} = \frac{y_{f}-y_{o}}{t_{f} - t_{o}} = \frac{7ft}{0.5s} = 14 ft/s (3)$