Question
(a)-Topic-4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration
(b)-Topic-4.3 Rates of Change in Applied Contexts Other Than Motion
(c)-Topic-4.6 Approximating Values of a Function Using Local Linearity and Linearization
(d)-Topic-4.3 Rates of Change in Applied Contexts Other Than Motion
2. Stephen swims back and forth along a straight path in a 50-meter-long pool for 90 seconds. Stephen’s velocity is modeled by \(v(t)=2.38e^{-0.02t}sin\left ( \frac{\pi }{56}t \right )\) where t is measured in seconds and v(t) is measured in meters per second.
(a) Find all times t in the interval 0 < t < 90 at which Stephen changes direction. Give a reason for your answer.
(b) Find Stephen’s acceleration at time t = 60 seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time t = 60 seconds? Give a reason for your
answer.
(c) Find the distance between Stephen’s position at time t = 20 seconds and his position at time t = 80 seconds. Show the setup for your calculations.
(d) Find the total distance Stephen swims over the time interval \(0\leq t\leq 90\) seconds. Show the setup for your calculations.
▶️Answer/Explanation
2(a) Find all times t in the interval 0 < t < 90 at which Stephen changes direction. Give a reason for your answer.
For 0 < t < 90, v (t ) = 0 ⇒ t = 56.
Stephen changes direction when his velocity changes sign. This occurs at t = 56 seconds.
2(b) Find Stephen’s acceleration at time t = 60 seconds. Show the setup for your calculations, and indicate units of measure. Is Stephen speeding up or slowing down at time t = 60 seconds? Give a reason for your answer.
v’ (60) = a(60) = −0.0360162
Stephen’s acceleration at time t = 60 seconds is −0.036 meter per second per second.
2(c) Find the distance between Stephen’s position at time t = 20 seconds and his position at time t = 80 seconds. Show the setup for your calculations.
\(\int_{20}^{80}v(t)dt\)
=23.383997
The distance between Stephen’s positions at t = 20 seconds and t = 80 seconds is 23.384 (or 23.383 ) meters.
2(d) Find the total distance Stephen swims over the time interval 0 ≤ t ≤ 90 seconds. Show the setup for your calculations.
\(\int_{0}^{90}\left | v(t)\right |dt\)
=62.164216
The total distance Stephen swims over the time interval 0 ≤ t ≤ 90 seconds is 62.164 meters.