Given the following, with the correct answer. What type of formula would be applied to this problem and how so?

An object falls freely in a straight line and experiences air resistance proportional to its​ speed; this means its acceleration is ​a(t)=−​kv(t), where k is a positive constant and v is the​ object's velocity. The speed of the object decreases from 1300 ft/s to 1200 ft/s over a distance of 1400 ft. Approximate the time required for this deceleration to occur.

I attempted to use the formula (1300^2 - 1200^2) / 1400^2 as suggested but it does not give the correct answer.

Correct Answer is: 1.1206

  • $\begingroup$ It sounds like it belongs on physics.se and we don't like to click through to find a problem, particularly a homework problem with no effort shown. $\endgroup$ – Ross Millikan Jan 28 '18 at 3:39
  • $\begingroup$ What does "divided by sfts/s" mean? $\endgroup$ – Mauve Jan 28 '18 at 3:44
  • $\begingroup$ This is a problem related to exponential decay/growth @RossMillikan so it should belong here. $\endgroup$ – bob Jan 28 '18 at 3:49
  • $\begingroup$ @Useless I fixed the problem. $\endgroup$ – bob Jan 28 '18 at 3:53

The units of your calculation are sec$^{-2}$ so that formula cannot be right. As a simple approximation, the average speed is about $1250$ ft/sec, so the deceleration time is $1400/1250=1.12$ seconds. For more accuracy, you should solve $s=\frac 12at^2+v_0t$ with your data which will lower the average speed slightly and increase the time slightly.

  • $\begingroup$ Thank you for the response and information, so if I were to set up the given information with this formula $s=\frac 12at^2+v_0t$ would it look something like this? s = 1/2(1250)(1.12) + (1.12) $\endgroup$ – bob Jan 28 '18 at 5:47
  • $\begingroup$ No You can't plug in $1.12-$ that is what we are trying to solve for. You also didn't square anything for the $\frac 12at^2$ term. We know $s=1400$. The velocity drops by $100$ so we have $at=-100$. Plug those in and solve for $t$. We get $1400=\frac 12 (-\frac {100}t)t^2+1300t$ $\endgroup$ – Ross Millikan Jan 28 '18 at 12:34
  • $\begingroup$ I solved for t which came out to 28/25 which is the same as the first approximation, is there possibly a more accurate method to getting this answer? @RossMillikan $\endgroup$ – bob Jan 28 '18 at 19:10

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