1. The swimmer changes direction at t = 2.5 s
2. At t = 5, the instantaneous velocity of the swimmer is 5 ft/s
3. The acceleration of the swimmer is 2 ft/s²
1. How to find where the swimmer changes direction?
Since the swimmers position is given by s(t) = t² - 5t + 6, the swimmer changes direction at the critical point, that is when ds/dt = 0.
So, ds/dt = d(t² - 5t + 6)/dt
= dt²/dt - d5t/dt + d6/dt
= 2t - 5
So, ds/dt = 0 implies that
2t - 5 = 0
t = 5/2
t = 2.5 s
So, the swimmer changes direction at t = 2.5 s
2. The instantaneous velocity of the swimmer at t = 5 s.
Since the position of the swimmer is s(t) = t² - 5t + 6, his velocity is v = ds/dt = d(t² - 5t + 6)/dt
= dt²/dt - d5t/dt + d6/dt
= 2t - 5
At t = 5
v = 2t - 5
v = 2(5) - 5
v = 10 - 5
v = 5 ft/s
So, at t = 5, the instantaneous velocity of the swimmer is 5 ft/s
3. The acceleration of the swimmer
The acceleration of the swimmer is a = dv/dt
= d(2t - 5)dt
= d2t/dt - d5/dt
= 2 - 0
= 2 ft/s²
So, the acceleration of the swimmer is 2 ft/s²
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