acceleration: constant dec. to stop in 45 ft from 30 mph

[kpx001]

the state of illinois cycle rider safety program requires motorcycle riders to be able to brake from 30 mph (44ft/sec) to 0 in 45 ft. what constant deceleration does it take to do that?

this problem im clueless but heres what i tried.

since im breaking i would get a(t) = -30dt .
-30t + c = 44 ?
c= -22/15
s(t) = -15t^2 - (22/15)t + c
45 = -15(0)^2 - (22/15)(0) + c .


am i doing this right so far?

 

[Deleted member 4993]

Re: acceleration word problem?

the state of illinois cycle rider safety program requires motorcycle riders to be able to brake from 30 mph (44ft/sec) to 0 in 45 ft. what constant deceleration does it take to do that?

this problem im clueless but heres what i tried.

since im breaking i would get a(t) = -30dt .<<< What is a(t)? 'a' generally represents (in these problems) 'acceleration.

-30t + c = 44 ? <<< How is that?

c= -22/15<<< How does this come from above?

s(t) = -15t^2 - (22/15)t + c <<< Where did this come from?

45 = -15(0)^2 - (22/15)(0) + c .


am i doing this right so far?

Let

t = time

a = acceleration

v = velocity

\(\displaystyle a = \frac{dv}{dt}\)

s = distance travelled

\(\displaystyle v = \frac{ds}{dt}\)

\(\displaystyle \frac{dv}{dt} \, = \, -a\)

\(\displaystyle \frac{dv}{ds} \cdot \frac{ds}{dt}\, = \, -a\)

\(\displaystyle v \cdot \frac{dv}{ds}\, = \, -a\)

\(\displaystyle v \cdot dv\, = \, -a\cdot ds\)

\(\displaystyle \frac{v^2}{2}\, = \, -a\cdot s + C\)

v = 44 when s = 0

C = 978

\(\displaystyle a \, = \, \frac{978 \, - \, \frac{v^2}{2}}{s}\)

You know v and s at the stopping point - find 'a'.

 

[tkhunny]

Re: acceleration word problem?

No,no. You need contstant decelleration and you don't know what it is. You started off already knowing!

a(t) = x <== That's ft/sec^2

v(t) = x*t + v0 <== That's ft/sec

s(t) = ½x*t^2 + v0*t + h0 <== That's ft

Now, what do we know?

v0 = 44 ft/sec
h0 = 0 ft

s(t) = ½x*t^2 + 44*t

And there must be reasonable values for x and t0 such that s(t0) <= 45

This is where this problem get's a little tricky. We have to think about it a bit from here unless you ave left out some of the problem statement. Without a time period in which to stop, there could be a few problems.

1) Quite obviously, there are limitations on stopping. If you hit a brick wall, you would stop very quickly, and certainly within 45 feet. Normally, however, this is not considered a valid stopping method. So, infinite decelleration is out of the question.

2) Suppose we were riding straight up and we just let gravity do the decelleration? This makes x = -32 ft/sec^2 and produces a 45 ft stopping time of ... What? We never get to 45 ft!! Well, okay, 1 g might be too much stopping power.

3) In order to stop EXACTLY on 45 ft, one sets s(t0) = 45 ft and solves for t0. This gives a lovely quadrtic equation that is easily solves and contains the discriminant 1936+90x. Since this must be > 0, we see that x > -968/45 = -21.511 and leads to about 2.045 seconds.

4) Of course, anything greater than -21.511 will work, but, like I said in #1, some decellerations will throw you off the bike.