Tricky constant acceleration question

[lavalamp]

This is an interesting question that I saw on Yahoo Answers, so I thought I'd share it. I hadn't seen a constant acceleration problem quite like this before, here it is, exactly as posted:

A rocket is launched straight up with constant acceleration. Three seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 8 seconds later. What is the speed of the bolt right at the moment when it falls off the rocket?

[S485122]

I will assume the earth as referential, ignore its rotation, ignore the decrease in the gravitational acceleration with altitude and considerations about air drag on the bolt, consider the speed of the bolt to be equal to that of the rocket at the moment if falls off (and ignore a lot of other complicating factors I have no doubt forgotten to think about.)

An object under constant acceleration "a" has at time "t" a speed of a * t + v[sub]i[/sub] where v[sub]i[/sub] is the initial speed. Its position is h[sub]t[/sub] = - 1 / 2 * a * t[sup]2[/sup] + v[sub]i[/sub] * t + h[sub]i[/sub] where h[sub]i[/sub] is it's initial position.

You know the position and speed of the rocket at t = 0 s : h = 0 m and v = 0 m/s.

Its hight after t seconds is 1 / 2 * a * t[sup]2[/sup], its speed (and the bolt's speed) is a * t where a is the rocket's acceleration.

After 3 seconds the rocket's hight will be : 9 / 2 * a and its speed will be : 3 * a.

You also have the following data : the bolt takes 8 seconds to fall back to earth moved by its gravitational acceleration "g" which is about -9,8 m/s[sup]2[/sup] (the minus sign is because gravitation decreases the hight.) The equation for its altitude is h = 1 / 2 * g * t[sup]2[/sup] + v[sub]i[/sub] * t + h[sub]i[/sub], where v[sub]i[/sub] and h[sub]i[/sub] are the initial speed and height : the same as the rocket's after 3 seconds.

Its altitude will be 0 after 8 seconds : 0 = 1 / 2 * g * 8[sup]2[/sup] + (3 * a) * 8 + (9 / 2 * a) => a = - 64 / 57 * g about 11 m/s[sup]2[/sup].

The speed of the rocket and thus of the bolt at the moment it falls of is thus about 33 m/s

Jacob

[lavalamp]

Originally Posted by S485122

I will assume the earth as referential, ignore its rotation, ignore the decrease in the gravitational acceleration with altitude and considerations about air drag on the bolt, consider the speed of the bolt to be equal to that of the rocket at the moment if falls off (and ignore a lot of other complicating factors I have no doubt forgotten to think about.)

Yeah, that's the usual case with problems like these.

I'm glad to see that our answers agree, that either means I didn't screw up, or we both screwed up the same way.

[cheesehead]

A rocket is launched straight up with constant acceleration. Three seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 8 seconds later. What is the speed of the bolt right at the moment when it falls off the rocket?

Wow -- brings back pleasant memories. That could be a typical Phys 1a midterm question circa 1967.

[cheesehead]

Originally Posted by S485122

Its position is h[sub]t[/sub] = - 1 / 2 * a * t[sup]2[/sup] + v[sub]i[/sub] * t + h[sub]i[/sub] where h[sub]i[/sub] is it's initial position.

The initial constant should be positive 1/2 there, instead of negative 1/2. It's correct later on.

[S485122]

Oops ! (Too much copy and paste after fiddling of line to have something more or less consistent. A failure thus :-(

Which form is 1a ? Which year of secondary school ? I could look it up but am too lazy.
;-)

Jacob

[cheesehead]

Originally Posted by S485122

Which form is 1a ? Which year of secondary school ? I could look it up but am too lazy.
;-)

That just happened to be the numbering for most introductory classes (a = first of three quarter terms) at the college I first attended. It's not some intercollegiate standard you could look up AFAIK.

[S485122]

Since you are from the USA I presume you mean the first year of higher education (after secondary school.) In France one learns those concepts in the last year of secondary school when following the scientific branch.

Jacob

[davieddy]

Originally Posted by lavalamp

This is an interesting question that I saw on Yahoo Answers, so I thought I'd share it. I hadn't seen a constant acceleration problem quite like this before, here it is, exactly as posted:

Very tricky:

Speed = v m/s
-3v/2 = 8v - 64g/2
v = 64*9.8/19 = 33

As you know, I got barred from Physicsforums for my "attitude"

David

[cheesehead]

Originally Posted by S485122

Since you are from the USA I presume you mean the first year of higher education (after secondary school.) In France one learns those concepts in the last year of secondary school when following the scientific branch.

Are you contrasting current French practice to my recollection of 1960s-era US practice? :)

My nephew was taught more advanced stuff in his circa-2000 US high-school (secondary school) chemistry class than I was in my 1960s US high-school chemistry class -- in fact, some things I wasn't taught even in my 1960s US post-secondary freshman/sophomore chemistry classes.

[PBMcL]

The problem as stated is missing an important piece of information, namely, what was the original height of the bolt at launch? So far everyone has assumed it fell off the bottom.

Suppose it's a very tall rocket and the bolt started on the side of the rocket at a height of (32 s^2) * g. Then a = 0 (the rocket is a dud) and v(t) = 0 is a valid solution.