A bug crawls back and forth on a number line with velocity given by the graph below. Positive velocities correspond to movement in the positive direction on the number line.

(a) If the bug is at 0 on the number line when *t* = 0, determine the position of the bug at *t* = 3, 5, and 8 seconds.

(b) If the bug is at -10 on the number line when *t* = 0, determine the position of the bug at *t* = 3, 5, and 8 seconds.

Answer

From *t* = 0 to *t* = 3 the bug travels

to the right.

From *t* = 3 to *t* = 5 the bug travels a weighted distance of

,

meaning it travels 3 units to the left.

From *t* = 5 to *t* = 8 the bug travels

to the right.

(a) At *t* = 0 the bug is at 0 on the numberline.

From *t* = 0 to *t* = 3 the bug moves 5 units right, so at *t* = 3 the bug is at 5.

From *t* = 3 to *t* = 5 the bug moves 3 units left, so at *t* = 5 the bug is at 2.

From* t* = 5 to *t* = 8 the bug moves 10 units right, so the bug ends up at 12

on the numberline when *t* = 8.

(b) The bug will be moving the same distances, but with a different starting place.

At time *t* = 0 the bug is at -10. From *t* = 0 to *t* = 3 the bug moves right 5, so when *t* = 3 the bug is at

-10 + 5 = -5.

From *t* = 3 to *t* = 5 the bug moves left 3, so when *t* = 5 the bug is at

-5 – 3 = -8.

From *t* = 5 to *t* = -8 the bug moves right 10, so when *t* = 8 the bug is at

-8 + 10 = 2.