Answer:
- yes
- v(t) gives the instantaneous velocity at time t
(a) correct
(b) -48 ft/s
(c) -32 ft/s; -64 ft/s
(d) 9.257 s
(e) -296.216 ft/s
Step-by-step explanation:
You are given an equation of motion and asked to find velocity and time at different points in the motion. This requires you differentiate the position function to find velocity, and that you solve the position function to find the time at a particular position.
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<h3>(a)</h3>
The coin is "dropped", so is presumed to have no initial velocity. The initial height is given as 1371 feet. Putting these values into the given equation, you have ...
s(t) = -16t² +1371
The velocity is the derivative of this function:
v(t) = -32t
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<h3>(b)</h3>
The average velocity on an interval is the change in position divided by the change in time:
(s(2) -s(1))/(2 -1) = ((-16·2² +1371) -(-16·1² +1371))/1 = -16(4 -1) = -48 . . . ft/s
(Note: the average velocity for a quadratic position function is the velocity at the midpoint of the interval: v(1.5) = -32(1.5) = -48 . . . ft/s.)
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<h3>(c)</h3>
Use the velocity function you found in part (a) to determine the instantaneous velocity at the given times:
v(1) = -32(1) = -32 . . . ft/s
v(2) = -32(2) = -64 . . . ft/s
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<h3>(d)</h3>
The coin will reach the ground when s(t) = 0. Solving for t, we find ...
s(t) = 0
-16t² +1371 = 0
1371 = 16t²
t = √(1371/16) = (√1371)/4 ≈ 9.257 . . . . seconds
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<h3>(e)</h3>
The velocity at impact will be ...
v(√1371/4) = -32(√1371/4) = -8√1371 ≈ -296.216 . . . ft/s
