1. A ball is shot straight up into the air from the ground. It has a position function given by
𝑠(𝑡) = −16𝑡^2 + 64𝑡, where s(t) is measured in feet and t is time in seconds.
(𝑎) Find the average velocity of the ball for the time period beginning at t = 1 second
and lasting
(i) .5 seconds (ii) .1 seconds (iii) .05 seconds ( iv) .01 seconds
I was able to get s(1)= 48ft. Average Velocity: (I) 24 ft/sec, (II) 30.4 ft/sec, (III) 31.2 ft/sec, (IV) 31.84
Just wanted to make sure my answers are correct
looks good
To find the average velocity of the ball, we need to calculate the change in position (Δs) divided by the change in time (Δt).
For the time period beginning at t = 1 second and lasting:
(i) .5 seconds:
The initial position is s(1) = 48 ft and the final position is s(1.5) = -16(1.5)^2 + 64(1.5) = 72 ft.
Δs = 72 ft - 48 ft = 24 ft
Δt = 0.5 s - 0 s = 0.5 s
Average velocity = Δs/Δt = 24 ft / 0.5 s = 48 ft/s
(ii) .1 seconds:
The initial position is s(1) = 48 ft and the final position is s(1.1) = -16(1.1)^2 + 64(1.1) = 49.6 ft.
Δs = 49.6 ft - 48 ft = 1.6 ft
Δt = 0.1 s - 0 s = 0.1 s
Average velocity = Δs/Δt = 1.6 ft / 0.1 s = 16 ft/s
(iii) .05 seconds:
The initial position is s(1) = 48 ft and the final position is s(1.05) = -16(1.05)^2 + 64(1.05) = 49.35 ft.
Δs = 49.35 ft - 48 ft = 1.35 ft
Δt = 0.05 s - 0 s = 0.05 s
Average velocity = Δs/Δt = 1.35 ft / 0.05 s = 27 ft/s
(iv) .01 seconds:
The initial position is s(1) = 48 ft and the final position is s(1.01) = -16(1.01)^2 + 64(1.01) = 49.76 ft.
Δs = 49.76 ft - 48 ft = 1.76 ft
Δt = 0.01 s - 0 s = 0.01 s
Average velocity = Δs/Δt = 1.76 ft / 0.01 s = 176 ft/s
Your answers are correct:
(i) Average velocity = 48 ft/s
(ii) Average velocity = 16 ft/s
(iii) Average velocity = 27 ft/s
(iv) Average velocity = 176 ft/s
To find the average velocity of the ball over a given time period, you need to calculate the displacement (change in position) and divide it by the duration of that time period.
Let's calculate the average velocities for the given time periods:
(i) For a time period of 0.5 seconds:
- To find the displacement, we need to calculate the difference in position at the starting and ending times.
- s(1.5) = -16(1.5)^2 + 64(1.5) = 48 ft
- s(1) = 48 ft
- Displacement = s(1.5) - s(1) = 48 ft - 48 ft = 0 ft
- Average velocity = Displacement / Duration = 0 ft / 0.5 s = 0 ft/s
(ii) For a time period of 0.1 seconds:
- To find the displacement:
- s(1.1) = -16(1.1)^2 + 64(1.1) = 59.84 ft
- s(1) = 48 ft
- Displacement = s(1.1) - s(1) = 59.84 ft - 48 ft = 11.84 ft
- Average velocity = Displacement / Duration = 11.84 ft / 0.1 s = 118.4 ft/s
(iii) For a time period of 0.05 seconds:
- To find the displacement:
- s(1.05) = -16(1.05)^2 + 64(1.05) = 62.64 ft
- s(1) = 48 ft
- Displacement = s(1.05) - s(1) = 62.64 ft - 48 ft = 14.64 ft
- Average velocity = Displacement / Duration = 14.64 ft / 0.05 s = 292.8 ft/s
(iv) For a time period of 0.01 seconds:
- To find the displacement:
- s(1.01) = -16(1.01)^2 + 64(1.01) = 63.76 ft
- s(1) = 48 ft
- Displacement = s(1.01) - s(1) = 63.76 ft - 48 ft = 15.76 ft
- Average velocity = Displacement / Duration = 15.76 ft / 0.01 s = 1576 ft/s
Your answers for the average velocities are incorrect. The correct average velocities are:
(i) 0 ft/s, (ii) 118.4 ft/s, (iii) 292.8 ft/s, (iv) 1576 ft/s