For an object whose position is x(t),
Distance traveled = ∫_a^b |v(t)| dt (the integral of the absolute value of velocity)
When a force F(x) is applied to an object from x = a to x = b,
Work W = ∫_a^b F(x) dx
Example ① The distance traveled from t = 0 to t = 3 by an object moving with velocity v(t) = t² − 4t + 3 m/s.
Solution: v = (t−1)(t−3)。On [0,1], v>0; on [1,3], v<0。
∫_0^1 v dt + ∫_1^3 (−v) dt = (1/3 − 2 + 3) + ((9−18+9) − (1/3−2+3))
= 4/3 + |0 − 4/3| = 4/3 + 4/3 = 8/3 m
Solution: v = (t−1)(t−3)。On [0,1], v>0; on [1,3], v<0。
∫_0^1 v dt + ∫_1^3 (−v) dt = (1/3 − 2 + 3) + ((9−18+9) − (1/3−2+3))
= 4/3 + |0 − 4/3| = 4/3 + 4/3 = 8/3 m
Key Point
- “Displacement” and “distance traveled” are different (distance traveled is the integral of the absolute value)
Practice Problems
- The distance traveled from t=0 to t=4 when v(t) = 2t − 6.
Solution and Explanation
- Answer: 10
On [0,3], v<0; on [3,4], v>0。∫_0^3(6−2t)dt + ∫_3^4(2t−6)dt = 9 + 1 = 10。