A particle is moving on a straight line in such a way that its velocity v is given by v(t)=2t+1 for 0≤t≤5 where t is measured in seconds and v in meters per second. What is the total distance traveled (in meters) by the particle between times t=0 seconds and t=5 seconds ?

Question

I was thinking I would do something like (2(0)+1) + (2(5)+1) but that didn't yield the correct answer which is 30.

Answer 1

if v(t) = 2t + 1
then s(t) = t^2 + t + c

s(5) = 25+5+c = 30+c
s)0) = c

effective distance = s(5) - s(0)
= 30+c - c
= 30

what you found is the difference in the velocities

Answer 2

Is there a way to do this without taking the integral of the equation which is what I believe you did? I'm just beginning integral calculus so we are still doing more basic methods.

Answer 3

To get distance from the velocity you must integrate.