A cat is climbing a tree. The cat is one foot above ground when its owner starts a stopwatch. The cat's velocity, in feet per second, is given by the following graph:

Answer

When *t* = 0 the cat is one foot above the ground:

(a) The area between the graph of *v*(*t*) and the *t*-axis on [0,3] is 6:

This means from time 0 to 3 seconds the cat climbed

Since the cat started 1 foot above the ground, when *t* = 3 it is 7 feet above the ground.

(b) The area between the graph of *v*(*t*) and the *t*-axis on [3,6] is 11:

This means from *t* = 3 to *t* = 6 the cat climbed

Since the cat was 7 feet above ground when *t* = 3, when *t* = 6 the cat is

7 + 11 = 18 feet above ground.

(c) The area between the graph of *v*(*t*) and the *t*-axis on [6,8] is 12:

This means from *t* = 6 to *t* = 8 the cat climbed

Since the cat was 18 feet above ground when *t* = 6, when *t* = 8 the cat is

18 + 12 = 30 = feet above ground.

(d) The area between the graph of *v*(*t*) and the *t*-axis on [8,10] is 12:

This means from *t* = 8 to *t* = 10 the cat climbed

Since the cat was 30 feet above ground when *t* = 8, when *t* = 10 the cat is

30 + 12 = 42 feet above ground.