# total distance traveled with calculus

**1. The problem statement, all variables and given/known data**

Suppose a particle responds to this equation of motion: s = t

^{2}– 5t + 6

a) find the velocity at two seconds and three seconds

b) find the acceleration

c) find the total distance covered after 3 seconds

**2. Relevant equations**

s = t^{2} – 5t + 6

v = ds/dt = 2t – 5

a = dv/dt = 2

**3. The attempt at a solution**

ai) 2t – 5 = 4 – 5 = -1

aii) 2t – 5 = 6 – 5 = 1

(everything is fine here)

b) a = 2

c) The equation given above is for position, since it uses velocity. Is it actually possible to find the distance covered from the displacement?

I also tried finding the position at t = 1, t = 2 and t = 3 to find the distance.

at t = 1, s = 2

at t = 2, s = 4 – 10 + 6 = 0

at t = 3, s = 9 – 15 + 6 = 0

How is it possible that the displacement is 0 twice consecutively if the velocity is non-zero? Isn’t that impossible?

I was actually going to calculate the displacement each time then add all together. The book had also already given an answer: 6.5m, although i’m not sure of how they got that.

Thanks for any answers.

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