How to get the velocity and the displacement from acceleration

Question

Suppose that a particle moves along a straight line with acceleration defined by a(t) = t − 3, where 0 ≤ t ≤ 6 (in meters per second per second). Find the velocity (in meters per second) and displacement (in meters) at time t if v(0) = 3 and d(0) = 0.

v(t)= ?

d(t)= ?

Find the total distance traveled (in meters) up to t = 6. (Round your answer to two decimal places.)

I don't know how to get the v(t) and d(t) from a(t), I know it has to do something with anti-derivatives, but I don't know how to apply it here.

Answer 1

yes for sure

v(t)=∫(t-3)dt=(t²/2)-3t+C

v(3)=0 calculates C=9/2

v(t)=(t²/2)-3t+9/2

note v(t) never negative, t=3 is the minimum for the quadratic (use quadratic formula)

dist(t)=∫((t²/2)-3t+9/2)dt=(t³/6)-(3t²/2)+(9t/2)+K

dist(0)=0 calculates K=0

dist(t)=(t³/6)-(3t²/2)+9t/2

when t =6s. distance=15 meters

Answer 2

a(t)= t-3 for first 6 seconds from 0<t<6

v(t)= integral of a(t)

d(t)= integral of v(t)

velocity = integral of a(t)=v(t) = t^2/2-3t + c, since v(3) = 0

9/2-9 +c= 0

c= 9-9/2 = 9/2

v(t) = t^2/2-3t + 9/2

d(t)= integral of v(t) = t^3/6-3t^2/2+9t/2+ k

d(0) =0

means d(0)= 0+k=0 and k=0

d(t) =t^3/6 -3t^2/2 + 9t/2

v(t) =t^2/2-3t+4.5

v' = t -3 = 0

t = 3 is when the particle changes direction and begins moving forward, instead of backwards

from t = 0 to t= 3 the particle moves backwards

from t=3 to t =6, it moves forward

sum the two distances to get total distance moved