Question

How do I find the instantaneous velocity of 0.55s?

Answer 1

We'll need some more info :)

Answer 2

Time (s) 0.0 0.20 0.30 0.40 0.50 0.60
Position (cm [W]) 0 25 75 75 75 0

The correct answer is 7.5 cm/s [E]

Answer 3

but I tried:

->
Vin = 37.5 cm
--------
5s
= 7.5 cm/s {?]

Answer 4

Well, from 0.50 to 0.60s you went from 75cm to 0. 75cm/10 = 7.5cm/s

Answer 5

Of course, that's Vavg during that time period...which isn't exactly what they're asking but if you know the answer is 7.5, then maybe the question is worded wrong

Answer 6

So what should I write?

Answer 7

75/10 = 7.5 cm/s

Answer 8

you got the 75 from?

Answer 9

0.50 s/

Answer 10

From your table. At 0.5s, the distance is 75cm

Answer 11

ohhh, okay. And if I find the instantaneous velocity at time 0.40s, the answer is 0.0 m/s why is that?

Answer 12

if you don't mind me asking.. D:

Answer 13

Because you can see from the table that the object is at the same position before and after 0.40s

Answer 14

So it is implied that it isn't moving at that point which = 0 cm/s

Answer 15

ohh, okay.

Answer 16

So when I find the time of 0.55s, it is basically 75 (because of 0.50s) - 0 (final velocity) / 10 s (because of .50 to 0.60s)?

Answer 17

Exactly

Answer 18

Also.. if I want to find the time of 0.10s I subtract 25 cm - 0 / 10?

Answer 19

yes

Answer 20

why so? :(

Answer 21

Ok, disregard what I just deleted :) Here's a better way to explain the 7.5 cm/s:

At 0.50s your position is 75cm which is 0.75m
At 0.60s your position is 0cm or 0 m

0.75/0.10 = 7.5cm/s

Answer 22

I shoulda remember to convert the distance to meters first...which is why it didn't look right at first. Anyway, the last message I posted is the correct way to demonstrate the answer.

Answer 23

ohh..

Answer 24

Sorry for the confusion...

Answer 25

It's okay. So first I convert it to cm = m?

Answer 26

Yes. I try to always convert things first to meters, seconds, kilograms, etc and when I don't, I get into trouble :)

Answer 27

ohhh, okay. But why?

Answer 28

All of the kinematic equations, force formulas, etc...are based on the standard units of meters, seconds, kilograms, etc.

Answer 29

ohh..

Answer 30

Here's some of the standard SI units which you should use (converting if you need to):

Length - meter
Mass - kilogram
Time - second
Electric current - ampere
Thermodynamic temperature - kelvin
Amount of substance - mole
Luminous intensity - candela

Answer 31

What if I want to find the instantaneous velocity at time 0.55 s, isn't that like the middle? of 0.50 - 0.60? Which will have the middle of 75 which is 37.5 cm?

Answer 32

one sec...this may help

Answer 33

I'm so sorry for bothering you D;

Answer 34

It's no bother at all :)
\[V_{inst}=\frac{\Delta d}{\Delta t}\]
Where d is displacement and t is time. The deltas are always calculated as the final value minus the initial value.

Anyway, what that means is that there's no need to worry about what's in the middle when calculating the delta.

So the difference in displacement between 0.50s and 0.60s is 0m - 0.75m = -0.75m.

The difference in time is 0.60s - 0.50s = 0.10s

Note that the distance is negative because you're essentially heading back to 0 (to the left) on the x-axis

Answer 35

err..in the last line I meant "the velocity is negative"...I'm all screwed up today

Answer 36

yup, I knew about the final and initial value.
Really? I didn't know that I didn't have to know what's in the middle when calculating the delta..
But how come the correct answer is 7.5 cm [E]?

Answer 37

If you actually use the formula you get:\[V_{inst}=\frac{-0.75m}{0.1s}=-7.5 m/s\]It's magnitude is 7.5 and it's direction is left (since it's negative)

Answer 38

yup. But I'm confused about the East since the data shows West.

Answer 39

and the answer is 7.5 cm/s

Answer 40

The delta is merely the difference between the 2 values...what's in the middle doesn't really matter.

In reality, you have no idea what's between those two values because you only know what the values are at those specific points. For all you know, the velocity have jumped up to 20 cm/s at 0.55s...but you have to go on what you do know.

Answer 41

ohh, yeah.

Answer 42

Ok, I didn't see the West part. That explains why your final answer is positive...which I thought was just because they only wanted the magnitude.

So the revised (and hopefully final equation) would be:
\[V_{inst}=\frac{(0m - (-0.75m)}{(0.60s - 0.50s)} = +7.5 m/s\]

Answer 43

why the -0.75 m?

Answer 44

I'm assuming West is negative:

Answer 45

because it's going down?

Answer 46

ohhh, yup.

Answer 47

I hope I havent confused you more by all this :) I have it straight in my head but it doesn't seem to be coming out very clear

Answer 48

oh my, it just feels overwhelming because there are so much info but thank you so much for trying to explain these concepts to me! I really appreciate them

Answer 49

You'll get the hang of it...it just takes some practice. In reality, you can do this problem without converting units but once you move on to harder problems, it's safer just to convert them first and then work the problem. That's my opinion anyway

Answer 50

yeah, I'll remember that! Thank you so much!

Answer 51

btw, when I find the instantaneous velocity of the time: 0.10s which has 25 cm [W], I subtract

Answer 52

brb...wife

Answer 53

it will be 0.25 m and 0-0.25 m / 10 s =

Answer 54

ahh, okay.

Answer 55

Sorry for the delay...one sec