1. **State the problem:** An object falls from rest with displacement given by $s = 16t^2$, where $s$ is the distance fallen in feet after $t$ seconds. We need to find: (a) The average speed during the first 5 seconds. (b) The instantaneous speed at the end of 5 seconds. 2. **Given:** $$s = 16t^2$$ 3. **Part (a): Average Speed during first 5 seconds** Average speed is total distance divided by total time: $$\text{Average speed} = \frac{s(5) - s(0)}{5-0}$$ Calculate displacement at $t=5$: $$s(5) = 16 \times 5^2 = 16 \times 25 = 400\text{ feet}$$ At $t=0$, $s(0) = 16 \times 0^2 = 0$. So, $$\text{Average speed} = \frac{400 - 0}{5} = \frac{400}{5} = 80\text{ feet per second}$$ 4. **Part (b): Instantaneous speed at $t=5$ seconds** Instantaneous speed is the magnitude of velocity, which is the derivative of displacement with respect to time: Calculate velocity: $$v(t) = \frac{ds}{dt} = \frac{d}{dt}(16t^2) = 32t$$ So at $t=5$: $$v(5) = 32 \times 5 = 160\text{ feet per second}$$ **Final answers:** (a) Average speed during first 5 seconds is $80$ feet per second. (b) Instantaneous speed at 5 seconds is $160$ feet per second.