# Total distance traveled with calculus

## Homework Statement

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

## Homework Equations

s = t2 - 5t + 6
v = ds/dt = 2t - 5
a = dv/dt = 2

## 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.

Related Introductory Physics Homework Help News on Phys.org
Look at the graph s=t^2 - 5t + 6

• 1 person
Gold Member
How is it possible that the displacement is 0 twice consecutively if the velocity is non-zero? Isn't that impossible?
The velocity is non-zero but it's direction changed

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.

NascentOxygen
Staff Emeritus
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