We get
vf = v 0 1 = 30
Divide both sides by 30: = e.
Take the log of both sides: ln = −.
Multiply both sides by − : − ln = t ≈ 113 seconds.
(b) How far, to the nearest 10 meters, will the body coast during the time it takes to slow from
30 m/s to 1 m/s?
Now we need to solve the differential equation = v 0 , which we can do with separation
of variables.First, multiply both sides by dt: ds = v 0 dt.
Integrate both sides: .
Evaluate the integrals: s = .
Now, plug in the initial conditions to solve for C: 0 = .
C = = 1,000
Therefore, . Now, we plug in the time t = 113 that we found in part
(a) as well as the initial conditions to solve for s, which yields the following: