# Total distance traveled with calculus

1. Jun 12, 2014

### albertrichardf

1. The problem statement, all variables and given/known data
Suppose a particle responds to this equation of motion: s = t2 - 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 = t2 - 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.

2. Jun 12, 2014

### lep11

Look at the graph s=t^2 - 5t + 6

3. Jun 12, 2014

The velocity is non-zero but it's direction changed

4. Jun 12, 2014

### dauto

Start from t=0 and keep in mind that the velocity is changing so the object will go one way, stop, and come back covering the same ground again explaining why the position at t=2 seconds coincides with the position at t = 3 seconds.

5. Jun 12, 2014

### Staff: Mentor

Yes, distance through whch the wheels have turned can be found by adding up a number of individual displacements. But you don't calculate those separate displacements by dividing the trip according to some arbitrary time interval chosen because of convenience to you. The necessary time intervals are dictated by some characteristic of the motion itself.