Calculus displacement question

Question

The velocity (in meters/sec) of a particle moving along a straight line is given by v(t)=2t^2+2t+1, where t is measured in seconds. Answer the following questions given that the initial position s(0)=1.

What is the meter position of the particle at any given time t?

What is the meter position of the body at time t=3?

What is the meter position of the particle at time t=10?

What is the displacement of the particle on the time interval 3≤t≤10?

Answer 1

v(t) = 2t² + 2t + 1

s(t) = ∫v(t)dt = (2/3)t³ + t² + t + C

Since s(0) = 1, C = 1.

So, s(t) = (2/3)t³ + t² + t + 1.

Find s(3), s(10), and s(10) - s(3).

Answer 2

v(t) = 2t^2 +2t + 1 = velocity of a particle moving in a line, in meters per second

position of the particle at any given time t is?

to find distance integrate v

s(t) - integral of 2t^2 +2t +1 = (2/3)t^2 + (2/3)t + t + c

s(0) = 1 = 2/3 +2/3 + c

c= 1-4/3 =-1/3

s(t) = (2/3)t^2 +(2/3)t + t - 1/3

at t=3, after 3 seconds s(3) = (2/3)3^2 +(2/3)3 + 3 -1/3 = 6+2+3 -1/3 = 10 2/3 meters

at t=10 after 10 seconds s(10) = (2/3)100 +(2/3)10 +10 -1/3 = 220/3 +10 -1/3 = 83 meters

displacement on the interval 3<t<10

is 83-10 2/3 = 72 1/3 meters change in position