Tutorial Exercise If an object is thrown upward at 128 feet per second from a height of 48 feet, its height S after t seconds is given by the following equation. S(t) = 48 + 128t - 16t (a) What is the average velocity in the first 4 seconds after it is thrown? (b) What is the average velocity in the next 4 seconds? Step 1 (a) What is the average velocity in the first 4 seconds after it is thrown? Recall that the average rate of change of a function y = f(x) from x = a to x = b is given by /b) - f(@) b - a We are given S(t) = 48 + 128t - 16t , where S is the height (in feet) of an object thrown upward after t seconds, and asked to find its average velocity in the first 4 seconds after it is thrown. Note that average velocity is a change in distance divided by a change in time and has units of distance per time v distance per time . So, the average rate of change formula can be adapted to find an average velocity, V, as follows. average rate of change = _(b) - (@ - S(t2) - S(t, ) b- a tz - $ 1 To find the average velocity of the object in the first 4 seconds after it is thrown, let t, = 0 and tz = 4 | 4. Step 2 Before we can find the average velocity, v. = _12 21 fort, = 0 to to = 4, we need to determine S(0) and 5(4). $2 - t , To find S(0), replace t with 0 and simplify. S(t) = 48 + 128t - 16t2 5(0) = 48 + 128(0) - 16 Now find S(4). 5(t) = 48 + 128t - 16t2 S(4) = 48 + 128 ) - 16(4)2