Can someone explain this? (Speed=Distance*Time)

[dsryan]

I'm in the process of joining the Police service and as a part of the entrance I'm working on my maths and seem to have hit a wall with speed / distance x time. Can't find much on the internet regarding this. Baring in mind I've been out of school for some time now, I've had to relearn everything.

speed = distance / time
distance = speed * time
time = distance / speed

Now, for example, what if the question was as follows;

A car has traveled 62 miles and the journey has taken 45 minutes.

Answer using this method: 62/45 = 1 r 17

How can I get the speed? What is the further step here? I understand this method works if you write it in hours, but it doesn't work in minutes. How can I make all of these methods work in minutes?

 

[Nono713]

I'm in the process of joining the Police service and as a part of the entrance I'm working on my maths and seem to have hit a wall with speed / distance x time. Can't find much on the internet regarding this. Baring in mind I've been out of school for some time now, I've had to relearn everything.

speed = distance / time
distance = speed * time
time = distance / speed

Now, for example, what if the question was as follows;

A car has traveled 62 miles and the journey has taken 45 minutes.

Answer using this method: 62/45 = 1 r 17

How can I get the speed? What is the further step here? I understand this method works if you write it in hours, but it doesn't work in minutes. How can I make all of these methods work in minutes?

The speed = distance / time formula conserves units, meaning that if your distance is in miles and your time is in minutes then your speed obtained by this formula is simply in miles per minute. In this case:

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{62 ~ \text{miles}}{45 ~ \text{minutes}} \approx 1.378 ~ \text{miles per minute}$$

To get the speed in miles per hour, simply convert your time to hours (in this case 0.75 hours since 45 minutes is three quarters of an hour) and apply the same formula with the new time to get the speed in miles per hour. You get:

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{62 ~ \text{miles}}{0.75 ~ \text{hours}} \approx 82.667 ~ \text{miles per hour}$$

Please note that 1.378 miles per minute and 82.667 miles per hour represent the same speed just expressed differently so there is no paradox here.

Similarly you can change the units of distance to convert this to kilometres per hour (if you know how many kilometres are in a mile, or how many miles are in a kilometre, in order to convert between them) and so on. As long as your units are consistent, the formula will check out.

 

[dsryan]

The speed = distance / time formula conserves units, meaning that if your distance is in miles and your time is in minutes then your speed obtained by this formula is simply in miles per minute. In this case:

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{62 ~ \text{miles}}{45 ~ \text{minutes}} \approx 1.378 ~ \text{miles per minute}$$

To get the speed in miles per hour, simply convert your time to hours (in this case 0.75 hours since 45 minutes is three quarters of an hour) and apply the same formula with the new time to get the speed in miles per hour. You get:

$$\text{speed} = \frac{\text{distance}}{\text{time}} = \frac{62 ~ \text{miles}}{0.75 ~ \text{hours}} \approx 82.667 ~ \text{miles per hour}$$

Please note that 1.378 miles per minute and 82.667 miles per hour represent the same speed just expressed differently so there is no paradox here.

Similarly you can change the units of distance to convert this to kilometres per hour (if you know how many kilometres are in a mile, or how many miles are in a kilometre, in order to convert between them) and so on. As long as your units are consistent, the formula will check out.

Thanks a lot for the reply! How would I get the sum to get the .75?

I'm trying to put together a method that I can always go back to in order to answer many other questions in the same manner, for example:

1). If the time taken is in minutes, do a sum to get the .75.
2). Do the formula.
3). Have the final answer.
Or, perhaps, if you could give me the sum as to how you got 1.378 then I could just times that by 60?

 

[Nono713]

Thanks a lot for the reply! How would I get the sum to get the .75?

I'm trying to put together a method that I can always go back to in order to answer many other questions in the same manner, for example:

1). If the time taken is in minutes, do a sum to get the .75.
2). Do the formula.
3). Have the final answer.
Or, perhaps, if you could give me the sum as to how you got 1.378 then I could just times that by 60?

I got 1.378 by dividing 62 by 45. Are you asking how to divide two numbers to multiple decimal places? Using remainders doesn't work too well with units, they are difficult to interpret (e.g. "1 remainder 17 out of 45 miles per minute"). Values like these are usually given as decimal approximations, here 1.378 miles per minute, or 1.4 if you want less precision.

And then, yes, there are 60 minutes in an hour so something in miles per minute for example can be converted to miles per hour by simply multiplying by 60.

 

[dsryan]

I got 1.378 by dividing 62 by 45. Are you asking how to divide two numbers to multiple decimal places? Using remainders doesn't work too well with units, they are difficult to interpret (e.g. "1 remainder 17 out of 45 miles per minute"). Values like these are usually given as decimal approximations, here 1.378 miles per minute, or 1.4 if you want less precision.

And then, yes, there are 60 minutes in an hour so something in miles per minute for example can be converted to miles per hour by simply multiplying by 60.

Ah, I see! So now I've identified my real problem.

I've never dealt with this before. Precision is key for the upcoming exam, so unfortunately for me I can't round it.

What would the sum look like to get the decimal of a remainder? For example 1 r 17?

Thanks again!

 

[Nono713]

Ah, I see! So now I've identified my real problem.

I've never dealt with this before. Precision is key for the upcoming exam, so unfortunately for me I can't round it.

What would the sum look like to get the decimal of a remainder? For example 1 r 17?

Thanks again!

If you know how to do long division, the process is quite simple, you just "keep going" after you reach the decimal point. For instance, see the second example in Long division - Wikipedia, the free encyclopedia. Basically just keep multiplying the remainder by 10 and repeating to get more and more digits. In your case it would go like:

62/45 = 1 remainder 17

(DECIMAL POINT)

Now multiply 17 by 10 to get 170, then 170/45 = 3 remainder 35

Now multiply 35 by 10 to get 350, then 350/45 = 7 remainder 35

Now multiply 35 by 10 to get 350, then 350/45 = 7 remainder 35

(it keeps going on forever)

So 62 / 45 is equal to 1.377777... which can be rounded to 1.378. If you ever reach a zero remainder, you're done so you stop and the decimal expansion simply ends, for instance 3/4 is 0.75. Sometimes it never ends, as in this case, but when dividing integers it always either eventually repeats or stops, so once you get the same remainder you already got before you can stop (because otherwise you'd just keep on going forever).

 

[dsryan]

see!

But, perhaps I'm missing something?, I can't seem to make it work for anything else.

A car has covered 97 miles in 192 minutes. What is the speed?

192/97 = 1 r 95

95 * 10 = 950

950/192 = 4 r 182

182 x 10 = 1820

1820 / 192 = 9 r 92

so, that would all together give me 1.49.

1.49 x 60 = 89.4

The correct answer should be 30.31 and not 89.4

What am I doing wrong?

Thanks again!

 

[MarkFL]

see!

But, perhaps I'm missing something?, I can't seem to make it work for anything else.

A car has covered 97 miles in 192 minutes. What is the speed?

192/97 = 1 r 95

95 * 10 = 950

950/192 = 4 r 182

182 x 10 = 1820

1820 / 192 = 9 r 92

so, that would all together give me 1.49.

1.49 x 60 = 89.4

The correct answer should be 30.31 and not 89.4

What am I doing wrong?

Thanks again!

\(\displaystyle v=\frac{d}{t}=\frac{97\text{ mi}}{192\text{ min}}\cdot\frac{60\text{ min}}{1\text{ hr}}=\frac{485}{16}\text{ mph}=30.3125\text{ mph}\)

 

[dsryan]

\(\displaystyle v=\frac{d}{t}=\frac{97\text{ mi}}{192\text{ min}}\cdot\frac{60\text{ min}}{1\text{ hr}}=\frac{485}{16}\text{ mph}=30.3125\text{ mph}\)

Hi Mark, thanks for the reply!

I'm sorry to sound really silly here but I didn't totally understand that. Could you explain it?

How are you getting the 485 and 16?Talk about maths for a dummy lol..

 

[MarkFL]

Hi Mark, thanks for the reply!

I'm sorry to sound really silly here but I didn't totally understand that. Could you explain it?

Talk about maths for a dummy lol..

What I did was the way my dad taught me to convert units when I was a kid. We know 60 minutes equal 1 hour, so if we multiply by the ratio of 60 minutes to 1 hour, we are multiplying by 1, and not changing the value. So, we begin with the relationship between speed $v$, distance $d$ and time $t$, solved for $v$, which is what we are asked to find:

\(\displaystyle v=\frac{d}{t}\)

Now, we are given $d=97\text{ mi}$ and $t=192\text{ min}$, so let's plug those into our formula:

\(\displaystyle v=\frac{97\text{ mi}}{192\text{ min}}\)

Now, we can see that our units here will be miles per minute, but we want to express the answer in miles per hour. So, if we can multiply this expression by something with minutes in the numerator, and hours in the denominator, then the minutes will cancel out, and we will have miles per hour. But, in order to not change the value of the expression, we must multiply by 1, so let's multiply by:

\(\displaystyle 1=\frac{60\text{ min}}{1\text{ hr}}\)

to get:

\(\displaystyle v=\frac{97\text{ mi}}{192\text{ min}}\cdot\frac{60\text{ min}}{1\text{ hr}}\)

Now, we may cancel the unwanted units:

\(\displaystyle v=\frac{97\text{ mi}}{192\cancel{\text{ min}}}\cdot\frac{60\cancel{\text{ min}}}{1\text{ hr}}\)

And we are left with:

\(\displaystyle v=\frac{97\cdot60}{192}\text{ mph}\)

Next we reduce:

\(\displaystyle v=\frac{97\cdot5\cdot\cancel{12}}{16\cdot\cancel{12}}\text{ mph}\)

to get:

\(\displaystyle v=\frac{97\cdot5}{16}\text{ mph}=\frac{485}{16}\text{ mph}=30.3125\text{ mph}\)