<p>Problem 1.
A pelican has a fish in his mouth. He's traveling 3 m/s. He drops fish. What is fish's distance from pelican after 2 s?
I did this
Vi = 3m/s for both the fish and the pelican
a = 9.8 (is this negative or pos?)
Then i used
x = Vi (T) 1/2 (a) (t) ^2
I got for this
25.6 m
Did I do this right? If no, what i am doing wrong..
Problem 2.
I am so confused on this. I have tried this for 65 minutes now.
Here it is.
A speeding car is traveling 92.0 km/h toward a police car at rest. facing the same direction as the speeding car. If the police car begins accelerating when the speeding car is 250.0 m behind the police car, what must the police car's acceleration be in order for the police car to reach the speeding car's velocity at the moment the speeding car catches up? Assume that the speed car does not slow down.</p>
<p>Ok i have no clue how to do this one. I tried everything
Can someone please give me detailed instructions on this one and tell me if my number 1 was correct?
Sorry if this is a burden</p>
<p>First convert the units to m/s, it's much easier.
So Lets cal the speeding car, car A, and police car car B. Then,</p>
<p>CAR A:
velocity = 92 km <em>1000 m / 60 min / 60 sec = 25.5 m/s
distance traveled as function of t: d(t)= 25.5</em>t</p>
<p>CAR B:
acceleration is unknown, but constant, so a(t) = a
velocity , v(t) = at
function of distance, d(t) = (at^2)/2 + 250 . 250 is 250 m head start</p>
<p>Both, the distance and the speed must be the same, so:
a<em>t = 25.5 , and, a= 25.5/t
25.5</em>t = (at^2)/2 + 250, so now substitute a = 25.5/t into the equation
25.5 * t = 25.5 <em>t /2 + 250 , and solve for t
when you solve for time, you get: t=19.6 sec
therefore a</em>t=25.5, and a =1.3 m/s^2</p>
<p>and your answer to the first problem seems to be correct, sorry that I can not check it for you, I am just not in the mental state to solve another physics problem.</p>
<p>can you explain this in like a little simpler terms.. i am kinda confused when you said function of distance, d(t) = (at^2)/2
Where are you getting at^2/2
Man thanks though
I hate this worksheet >.>.</p>
<p>it's from calculus, I simply take the antiderivative. Because acceleration is the rate of change in velocity, and the velocity is the rate of change of displacement, they are the derivatives of each other. This comes from calculus, and if you havent taken all you do is simply raise the power of the variable and divide by the new power.
So, if acceleation is a,
then velocity is v=a<em>t + constant(the initial velocity)
and distance is x=(a</em>t^2 )/2 + constant(the initial displacement)
You see, I simply integrate with respect to time, and each time I add a power of t, and divide by the new power, and the constant is simply the initial condition.</p>
<p>Woah lol.. i know he did not expect us to know that. It would look wierd if i put that. He only has told us the 3 equations (you know VF = VI + at and the other 2). Can you explain them to me with only those 3?
OH and also V = x/t.</p>
<p>The problem is impossible (maybe) to solve without that eqaution because you need the relationship to exist between time and distance, because you need to equate the two distances and the two velocities.
If I remember correctly the equation x= (a<em>t^2)/2 + v</em>t + d is also part of the three basic equations along with:
v=a<em>t + vi
and vf^2 = 2ax + vi^2
Without using the equation x= (a</em>t^2)/2 + v*t + d, I simply do not know how to solve the problem.</p>
<p>By the way you can derive x= (a<em>t^2)/2 + v</em>t + d by integrating VF = VI + at, they are the same thing, only one is derivative of the other. Just tell your teacher you know some calculus and you derived x= (a<em>t^2)/2 + v</em>t + d by use of common sense, he will not be able to contradict your statement.</p>
<p>Sorry have sleep now, you have the answer before your eyes, and you used my equation before yourself in the first problem and didnt realize it. Everything is solved and legitimate, good luck!</p>
<p>I got something way different now... here this
So I also got 25.556 for V for the S. Car
Now i solved for T of the S. Car and got 9.78
So try to understand this, the S. Car has to travel distance 250 + X where X is the distance it takes the police car to catch up to the S. CaR.
Also 9.78 + T equals the time the S. Car has to take, where T represents the time it takes for the police car to catch up to the S. Car
Now I did this
Knowing 25.556 = (250)/(9.78)
And that his speed does NOT change i just changed it to this
25.556 = (250 + x)/(9.78 + t)
So the ratio of X / T = 250/9.78 as well = 25.556
Now I made
X = 25.5556 T by cross mult.
And then I put the X in the equation for the distance travled by the police car
25.5556 T = 0 (T) + 1/2 (at^2)
So u end up with
51.112T = at^2
divide through by T and you get
51.112 = AT
now plug in 2nd equation Vf = Vi + at
And you get VF = 51.112.
Is this correct?</p>
<p>So you found the final velocity of police car to be 51.1 m/s? Well, according to the problem its speed must also equal, so it must be 25.5 m/s. Therefore your entire approach is wrong and overly complicated. You got 9.78 seconds for the speeding car to travel 250 m. That is useless, since the police car is at the same time accelerating, so the distance at which they meet is actually at 500m, and not 200. Dont try to work around the method I used, it is hopeless since that is the only(or at least it seems to me)proper solution.</p>
<p>You're right. The velocity's must be the same. The teacher gave us a day and I asked him and I read the question wrong.. the WHOLE TIME. I was sovling for VF and not A....</p>