A jogger jobs around a circular track with a diameter of 880 meters in 40 minutes. What is the distance around the track?

You need to know that the perimeter (or in this case, circumference) is given by:

Perimeter = 2\pi r

but since diameter = 2 radius,

Perimeter = 2\pi \frac{d}{2} = \pi d

Can you post the answer you get?

(\pi is approximately 3.14)

19.7??

Hmmm...

Perimeter = 3.14 x 880 = 2764.6 m

or you didn't mention something about the time given?

The whole problem itself is this...

A jogger jobs around a circular track with a diameter of 880 meters in
40 minutes.
(a) What was the time in seconds?
40 X 60=2400

(b) What is the distance around the track?

(c) What was the runner's average speed?

(d) What was the runner's average velocity?

The rest I cannot figure it out

I have no idea how to work out any of the problems! (really bad at math sorry)

That's okay, we're here to learn right? :)

Okay, the circumference is indeed 2764.6 m.

You need to multiply \pi (which is 3.14) by the diameter of the path, that is 880 and that's it!

Now for the next part, speed is given by:

Speed = \frac{Distance}{Time\ Taken}

Can you work this out now? :)

1.1519?

Also, I had followed your steps above and I had gotten 2763.2 may I ask if you rounded?

I actually took the complete value of \pi which is 3.1415927. I don't know what your book/question asks to use as when it comes to the value of pi. But if you got 2763.2, that's good too but you must say that you took pi as being 3.14 like this:

Taking \pi as 3.14,

Distance of track = 3.14 x 880 = 2763.2 m

Now, there is something called significant figures. Usually, you put your answer in the same amount of significant figures (sf) as in the given problem, or one sf better. In this one, the highest sf is 2 because of '880'. Hence, you can put your answer this way:


Distance of track = 3.14 x 880 = 2763.2 m = 2760 m (3 sf)

But that doesn't matter much and let's continue. I'll take your value for the distance.

~~~~~~~~~
Speed = 2763.2 / 2400 = 1.15 (3 sf)

Ok, good! :)

Last part now. Velocity is defined as the rate of change of displacement, that is:

Velocity = \frac{Displacement}{Time\ Taken}

Displacement now is the distance from the starting point to the finishing point. Can you find this distance and post the velocity? :)

I some how got 0 and I do not even know how I did that. But, I had thought I had a different answer and so I started typing. Well, the first number I put in was 0 and all the sudden a green check mark pops up saying it was correct? I don't know what I just did. (haha)

That's correct! :)

Yes, the velocity is really 0. This is because the jogger is back at his starting point right? It is a circular path, he ran around it and came back to the starting point. So the displacement becomes zero. No matter by what you divide zero, you'll get zero again, so, the velocity is zero!

Well done! :)

Awsome! Thank you so much you don't know how stressed I was because my teacher just was not explaining it like you just have! He didn't sit down and explain anything! I really truly appreciate it =]

I'm glad to help :)

If you have more problems, you can ask! :)

If you don't mind? I am stuck on this last problem of my homework, just like the problem above after 2 days, still have not yet figured it out!

A space probe of the surface of Mars sends a radio signal back to the
Earth,a distance of 8.86 x 1010 km. Radio waves travel at the
speed of light (3.00 x 108 m/s).


(a) How far away is the space probe?


(b) How many seconds does it take for the signal to reach the Earth?


(c) How many hours is this?

a) The distance from Mars to Earth is how much? Where is the space probe?

b) Use the formula for speed that I gave you earlier.

c) The time you got is in seconds. Convert this in hours.

Can you give it a try? :)

Yes I will try and if you don't mind respond back to make sure I did it right?

Yes, given I see your response in time. I just took lunch, and this is why I took some time in replying you back :o

For "A" I got 2.658?

Let's take it this way.

My computer is besides me. I am 2400 km away from you. How far is my computer from you?

2.4ft?
I believe I would divide 2400 into a 1000?

I am so confused please bare with me

You are used to the British system?

Ok, let's change this.

Um... My computer is besides me. I am 2400 miles away from you. How far is my computer from you?

2.2?

Don't you think that it's 2400 miles away from you? Do you get it or, do you still have something somewhere you're not sure about?

I just completely don't get it. All the book really says is to change km to m? I don't know how to do that and when I answer the question I have to put in Type in scientific notation, 4 x 1013 would be typed in as 4e13 like this? Well I don't how to get the answer from 8.86 X 10e10

In fact the answer is 8.86 \times 10^{10}\ km.

Let's take a smaller distance. You are in your house. The market is 10 yards away from your house. How far are you from the market, in yards?

I know every 1 yard is 3 feet so 10 x 3=30ft
so if you want yards wouldn't it be 10?

Right! But you didn't have to calculate anything here! The answer was already given to you! Do you see now? :)

Yes! So for "A" I did multiply 10 X 8.86 = 88.6 now the total answer is 88.6e12... I just honestly don't know how 12 is even in there or how I got it because I just figured 10e10 so I just started putting numbers down till I got the right answer

Ok, now to the question.

The distance of the probe is 8.86e10 km from the Earth.

Recall that 1 km = 1000 m.

So, in metres, 8.86e10 km = 8.86e10 x 1000 = 8.86e13 m

This is using scientific notation. It's just like:

8.86 \times 10^{10}\ \times\ 1000 = 8.86 \times 10^{10} \times 10^3 = 8.86 \times 10^{(10+3)} = 8.86 \times 10^{13}

Now, we come to the time it takes. Can you find it?

Speed = \frac{Distance}{Time\ Taken}

So,

Time\ Taken = \frac{Distance}{Speed}

So would I set up this problem like this time taken=88.6/3.00=29.5
or time take=88.6/1000

Time \ Taken = \frac{Distance}{Speed}

So, we have:

Time\ Taken = \frac{8.86\times10^{13}}{3\times 10^8}

You need to put everything in the number.

Maybe the scientific notation is not clear...

8.86\times 10^{13} = 88600000000000

3 \times 10^8 = 300000000

Getting back,

8.86\times 10^{10} km = 88600000000 km

88600000000 km = 88600000000 \times 1000 m = 88600000000000 m

Im sorry, I still don't get it. I am sorry if I am frustrating you =/

It's okay. It's just that I am thinking about how to go and make you understand... maybe this way...

Find:
I) 2 x 10 = ?
ii) 58 x 100 = ?
iii) 6.5 x 10000 = ?
iv) 3 x 10e2 = ?

20
5800
65000
300

For the last one, no.

3 x 10e2 = 3 x 10 x 100 = 3000

okay?

If this was: 3 x 1e2, then this becomes 3 x 1 x 100 = 300

OK

Ok, now this one:

3e5 x 6e2 = ?

300,000 x 600= 180,000,000

Correct. Now, you'll see something too.

180,000,000 in scientific notation is 1.8 x 10^8 , or 1.8e8

What happened?

3e5 x 6e2 = (3 x 6) x e(5 + 2) = 18e7

Now, this is 18 x 10^7 right?

or

(1.8 x 10) x 10^7

This becomes 1.8 x 10^8

Try these now, using this new way of multiplying numbers, giving your answer in scientific notation.

1e16 x 8e18 = ?

5e12 x 3e8 = ?

9e7 x 7e6 = ?

I am starting to understand so would I use these tools to figure out "B"

In a way...

For b)

Time taken = 8.86e13 / 3e8 = (8.86 / 3) x e(13-8)

OK

I finally was able to figure it out and it is 88600000000000 X 300000000=295333

Yes, that's correct, though you put the multiply sign in your post instead of the '/' sign ;)

Well done! :)

Why thank you! It was incredibly obvious and so I don'y know why I didn't get it.

If you don'y mind I got more hw and there are these 2 problems that I have not gotten.. do you mind helping me again please? =]

A student drops an object out the window of the top floor of a
High-rise dormitory.
(a) Neglecting air resistance, how fast is the object traveling when it
Strikes the ground at the end of 4.6 s?
(b) Express the answer to part a in miles per hour.
(c) How far, in meters, does the object fall in 4.6 s?

Now I know "a" is 45.09

But everything else is like.. Huh?

Then there is this problem..


Change in speed over change in time is acceleration...
A car travels on a straight, level road. Starting from rest, the car is
Going 30 m/s at the end of 3.0 sec.
(a) What is the acceleration of the car during the first 3.0
Seconds of motion?
In 9.0 more seconds, the car is going 40 m/s.
(b) What is the car's acceleration for this period
(during the 9 seconds)
The car then slows to 18 m/s in 2.0 seconds.
(c) What is the acceleration of the car during this period
Of time?
(d) What is the overall average acceleration of the car for the
Total time?

(a) Ok, using v = u + at,

v = 0 + 9.8(4.6) = 45.08 m/s

Seemed you made a typo... if you took a = 9.81, you'd get v = 9.81(4.6) = 45.126 m/s

(b) 1 mile = 1 609 metres
45.08 m = (1/1609) x 45.08 = 0.0280 miles

1 h = 3600 s

So, 45.08 m/s = 0.0280 mil/s

In 1 second, it does 0.0280 miles
In 3600 seconds, it does about (0.028 x 3600) = 101 miles.

Hence, speed = 101 miles/hr

(c) Use the formula:

s = ut + \frac12 at^2

u is the initial speed which is 0
t is the time which is 4.6 s
a is the acceleration due to gravity which is 9.8 m/s^2
s is the displacement, in this case, also the distance.
~~~~~~~~~~

Use the definition that you just gave me.

(a) Acceleration = \frac{[Final\ speed] - [Initial\ speed]}{[Final\ time] - [Initial\ time]}

Final speed is 30 m/s.
Initial speed is zero since the car starts from rest.
Final time is 3 s.
Initial time is zero.

(b) Use again the same thing:
Acceleration = \frac{[Final\ speed] - [Initial\ speed]}{[Final\ time] - [Initial\ time]}

Final speed is 40 m/s.
Initial speed is 30 m/s.
Final time is 12 s. (in all, 9 + 3 = 12 seconds)
Initial time is 3 s. (during the 9 seconds means it starts at 3 seconds)

(c) Use the same thing as above and try to calculate the new times.

(d) It's the same thing. Just take the final and initial speeds at the last instant and when the time was started.

Post what you get! :)

So for C I got -11

Right, the acceleration of the car is indeed -11 m/s^2 :)