s is the position of a particle traveling on a line at time, t, (in seconds) where s=t-1. find the position exactly at t=4 seconds in ft and find the velocity exactly at t=4seconds
3 and 3/4
wouldn't velocity be constant at 1?
Data: s = t - 1 [ft] t -> time [s] Searching: Position at t = 4 s; in ft; Velocity [ft/s] at t = 4 s; Resolution: s = 4 - 1 s = 3; v derivate of s in time: v = 1 [ft/s]
if velocity was constant at 1ft/sec, then at t = 1sec , the distance traveled should have been 1 foot.
i am not sure if I am right or the rest of you guys are right.
How are you getting 3/4
v = distance/time at t=4, s =3 so v = 3/4
I'm pretty sure that my answer is right. Look: The velocity is one feet per second. So in four seconds it wals four feet's. Them looking to the space equation you can see that the space is given by the distance walke minu one. What gives us three feets. At least i can't find any error in that.
Yeah, it already states the velocity.
the particle could be accelerating or traveling at the same rate both answers are correct.
If this is not a calc class then you are right
in my case it is accelerating, in your case it is constant
It's a first order equation. So the velocity is constant. There's no aceleration.
yes. I am giving only the discrete values of velocity at discrete intervals. that is why it is a first order equation.
the problem can be modeled both ways a) as a particle traveling at constant speed b) as a particle accelerating with decay and eventually reaching 1.
Sorry, but i can't see the aceleration in the modeling way (b).
Well, don't know ... Wish my words could help in some way.
Use your imagination a bit :)
position function=P P(x)=2x+1 What is function really mean is an object start at 1 and move with the speed of 2 So we are asked to find position at some time (say 3) based on velocity V=2, we would multiply that 2*3=6 but recall initial position is at 1 so 6+1=7
The initial position does not contribute to velocity so brunosantana is absolutely right.
your premise is incorrect. position at some time t depends on the value of the velocity at that time.
Yes but that velocity is constant, you are factoring initial position to find velocity
no, I am modeling velocity at discrete intervals given time and displacement.
you are presuming velocity is constant, I am not. both of us are correct, though.
So if I am running on 26 miles track but I start at 1st mile and run 2 mph what would my speed be?
again, you are assuming that you are running at a constant speed. I am not. the 2mph means that your speed is constant.
you already gave your answer in your question. I am saying there are two ways to model that equation.
Yes, to make this point:So by your method I am accelerating even when I am not?
no. you are assuming that you start at mile 1. in my method you start at mile 0 and accelerate slowly until you mimic what you get by running at a constant speed.
if you were traveling at 60 miles an hour for one hour and I traveled at 360 miles an hour for 10 minutes, the distance covered would be the same.
I still stand by points but good debate
I am not denying that your way is correct. I am just saying the same thing can be modeled two ways. good debate though :)
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