Question

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

Answer 1

3?

Answer 2

Yes 3

Answer 3

3 and 3/4

Answer 4

wouldn't velocity be constant at 1?

Answer 5

not necessarily.

Answer 6

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]

Answer 7

if velocity was constant at 1ft/sec, then at t = 1sec , the distance traveled should have been 1 foot.

Answer 8

i am not sure if I am right or the rest of you guys are right.

Answer 9

How are you getting 3/4

Answer 10

v = distance/time
at t=4, s =3
so v = 3/4

Answer 11

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.

Answer 12

Yeah, it already states the velocity.

Answer 13

the particle could be accelerating or traveling at the same rate
both answers are correct.

Answer 14

If this is not a calc class then you are right

Answer 15

in my case it is accelerating, in your case it is constant

Answer 16

It's a first order equation. So the velocity is constant. There's no aceleration.

Answer 17

yes. I am giving only the discrete values of velocity at discrete intervals. that is why it is a first order equation.

Answer 18

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.

Answer 19

Sorry, but i can't see the aceleration in the modeling way (b).

Answer 20

Well, don't know ... Wish my words could help in some way.

Answer 21

Use your imagination a bit :)

Answer 22

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

Answer 23

The initial position does not contribute to velocity
so brunosantana is absolutely right.

Answer 24

your premise is incorrect.
position at some time t depends on the value of the velocity at that time.

Answer 25

Yes but that velocity is constant, you are factoring initial position to find velocity

Answer 26

no, I am modeling velocity at discrete intervals given time and displacement.

Answer 27

you are presuming velocity is constant, I am not.
both of us are correct, though.

Answer 28

So if I am running on 26 miles track but I start at 1st mile and run 2 mph what would my speed be?

Answer 29

again, you are assuming that you are running at a constant speed. I am not.
the 2mph means that your speed is constant.

Answer 30

you already gave your answer in your question.
I am saying there are two ways to model that equation.

Answer 31

Yes, to make this point:So by your method I am accelerating even when I am not?

Answer 32

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.

Answer 33

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.

Answer 34

I still stand by points but good debate

Answer 35

I am not denying that your way is correct. I am just saying the same thing can be modeled two ways. good debate though :)