20∫(from t=0 to t=10)[1−e-0.5t]dt equals 20[t + 2e-0.5t|(from t=0 to t=10)]
which gives 20[10 + 2e-5−0−2] or 160.2695179 Meters.
Krystle I.
asked 04/21/20Hi, I am working on a problem and I think I have it solved, but would just like to make sure that I have done it correctly, if possible. I appreciate anyone taking the time to help me.
Problem:
"The velocity of a falling object may be modeled by v(t) = 20(1-e-0.5t) meters per second for the first ten seconds. Calculate the distance the object has fallen in the first ten seconds of the fall."
My Solution:
∫ 20(1-e-0.5t)dt = 20∫1dt - 20∫e-t/2dt
solving for first integral: ∫1dt = t
solving for second integral (u = - t/2, -1/2 = du/dt, -2du = dt): ∫e-t/2dt => -2∫eudu = -2eu = -2e-t/2
putting it all together: 20∫1dt - 20∫e-t/2dt = 40e-t/2+20t + C
Then evaluate the definite integral: 40e-10/2 + 20(10) - [ 40e-0/2 + 20(0)] ≈ 160.3 meters
Again, thank you for the help, ahead of time.
20∫(from t=0 to t=10)[1−e-0.5t]dt equals 20[t + 2e-0.5t|(from t=0 to t=10)]
which gives 20[10 + 2e-5−0−2] or 160.2695179 Meters.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.