The acceleration function for a particle moving along a line

Question

The acceleration function for a particle moving along a line is a(t)=2t+2a(t)=2t+2 . The initial velocity is v(0)=−8 Then:

a)The velocity at time tt, v(t)=v(t)=  

b)The distance traveled during the time interval [0,4] is equal to =  

Answer 1

a(t) = 2t +2

v(t) = integral of a(t) = t^2 +2t + c

v(0) = -8=c

v(t) = t^2 +2t -8

s(t) = integral of v(t) = t^3/3 + t^2 -8t + k

evaluate from 0 to 4

s(4)-s(0) = 64/3 +16 -32 -0 = 64/3 -48/3 = 16/3 = 5 1/3= distance traveled from time t=0 to t=4

However, "distance traveled" is an ambiguous term. It could be equivalent to displacement, the distance from intial time to 4 seconds later. Or it could mean any movement in either direction back & forth, all summed together. adding the absolute value of any negative movements plus positive movements

s(2)-s(0)= |8/3 +12/3 - 48/3| = |8/3-36/3| =|-28/3| = -9 1/3 for the 1st two seconds, 9 1/3 backwards

then

s(4)= 5 1/3 - s(2) = 5 1/3 - - 9 1/3 = 14 2/3 distance traveled forward

14 2/3 + 9 1/3 = 24 = total distance traveled

the particle moved backwards 9 1/3, then moved forward 14 2/3, sum to 24

sort of like saying you were given $9, then you gave $14 back. where does that leave you? $5 behind? or recipient of exchanges totalling $24? sort of like a bumble bee doing a dance back and forth, going 5 1/3 feet, yet if you added up all its back and forth dances it sums to 24 feet