Question

A particle moves along the x axis with velocity of v(t) m/s. Find the displacement and distance travelled over the given time interval for:

a) v(t) = 3t-2 for 0 ≤ t ≤ 2
b) v(t) = |1-2t| for 0 ≤ t ≤ 2

Answer 1

velocity is the derivative of distance
so distance is the integral of velocity
integral of (3t-2) = (3/2) t^2 - 2t
d(t)=(3/2)t^2 -2t
t=0 -> d(0)=0
t=1 -> d(1)=3/2 -2 = -1/2
t=2 -> d(2)=(3/2)(4) - 2(2) = 6-4 = 2

Answer 2

so displacement is 2 m and distanc is 8/3 m.

Answer 3

yes the displacement is 2, but i think the distance isn't 8/3.
from t=0 to t=1 , the distance is 1/2
from t=1 to t=2, the distance is 2-(-1/2) = 3
so the total distance from t=0 to t=2 is 3 + 1/2 = 7/2

Answer 4

at t=3/2 v=0 the particle changes its direction of motion , u need to consider that also while finding the distance
hint:try plotting it on the x axis