(b)
(a)
Using v(1) = 6, we get 6 = 2(1)^2 − 3(1) + C 1 , and C 1 = 7, from which it
follows that v(t) = 2t^2 − 3t + 7. Since
Using f (2) = 5, we get ,
so . Thus,
.
When .For more examples of motion along a line, see Chapter 8, Further Applications
of Integration, and Chapter 9, Differential Equations.