The velocity of an object is v(t) = 32t ft/s.?

Question

Use Eq. (2) and geometry to find the distance traveled over the time intervals [0,2] and [2,5].

Eq. (2) is: Distance traveled during [t1, t2] = area under the graph of velocity over [t1, t2]

Answer 1

distance = [32t] evaluated from 0 to 2

= 16(2)^2 - 0 =64 feet

= [32t] evaluated from 2 to 5

= 16(5)^2 - 16(2)^2

= 336 feet

Answer 2

You need to integrate v(t) from t1 to t2

So over [0,2] it's the integral of 32t wrt t from 0 to 2 is 16t^2 evaluated at 0 and 2 = 64

Same process for [2,5]