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Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
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.
Given a velocity function $v\left(t\right)=3t-5$ (in meters per second) for a particle in motion from time $t=0$ to time $t=3,$ find the net displacement of the particle.
Applying the net change theorem, we have
The net displacement is $-\frac{3}{2}$ m ( [link] ).
Use [link] to find the total distance traveled by a particle according to the velocity function $v\left(t\right)=3t-5$ m/sec over a time interval $\left[0,3\right].$
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t -intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for t . Thus,
The two subintervals are $\left[0,\frac{5}{3}\right]$ and $\left[\frac{5}{3},3\right].$ To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval $\left[0,\frac{5}{3}\right],$ we have $\left|v\left(t\right)\right|=\text{\u2212}v\left(t\right)$ over that interval. Over $\left[\frac{5}{3},3\right],$ the function is positive, so $\left|v(t)\right|=v(t).$ Thus, we have
So, the total distance traveled is $\frac{14}{6}$ m.
Find the net displacement and total distance traveled in meters given the velocity function $f\left(t\right)=\frac{1}{2}{e}^{t}-2$ over the interval $\left[0,2\right].$
Net displacement: $\frac{{e}^{2}-9}{2}\approx -0.8055\phantom{\rule{0.2em}{0ex}}\text{m;}$ total distance traveled: $4\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}4-7.5+\frac{{e}^{2}}{2}\approx 1.740$ m
The net change theorem can be applied to the flow and consumption of fluids, as shown in [link] .
If the motor on a motorboat is started at $t=0$ and the boat consumes gasoline at the rate of $5-{t}^{3}$ gal/hr, how much gasoline is used in the first 2 hours?
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval $\left[0,2\right].$ We have
Thus, the motorboat uses 6 gal of gas in 2 hours.
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