568
Chapter 5 | Integration
40 dt + ⌠ ⌡ 4 5
2 5 | v ( t ) | dt = ∫ 2 4
∫
30 dt
=80+30 =110. Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function. Example 5.24 Finding Net Displacement
Given a velocity function v ( t ) =3 t −5 (in meters per second) for a particle in motion from time t =0 to time t =3, find the net displacement of the particle.
Solution Applying the net change theorem, we have ∫ 0 3
t |
3
(3 t −5) dt = 3 t 2
2 −5
0
⎡ ⎣ 3(3) 2
⎤ ⎦ −0
(3)
=
2 −5
= 27 2 −15 = 27 2 − 30 2 = − 3 2 .
The net displacement is − 3 2
m ( Figure 5.33 ).
This OpenStax book is available for free at http://cnx.org/content/col11964/1.12
Made with FlippingBook - professional solution for displaying marketing and sales documents online