# Time required for deceleration to occur

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

• 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. – Ross Millikan Jan 28 '18 at 3:39
• What does "divided by sfts/s" mean? – Mauve Jan 28 '18 at 3:44
• This is a problem related to exponential decay/growth @RossMillikan so it should belong here. – bob Jan 28 '18 at 3:49
• @Useless I fixed the problem. – 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.
• 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) – bob Jan 28 '18 at 5:47
• 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 $1400=\frac 12 (-\frac {100}t)t^2+1300t$ – Ross Millikan Jan 28 '18 at 12:34