Speed and distance maths question not physics

[MathsFormula]

This is a maths algebra question and not physics so I don't think I'm meant to know the formula speed = distance / time so I won't use it.

Question says : A man runs to a
telephone and back in 900 seconds. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone.

My method :

900 seconds is divided in the ratio 5:4

= 500 : 400

So has been travelling 400 seconds to the telephone and 500 seconds on the way back.

Therefore distance to the telephone is 400 seconds travelled at 5 m/s = 400 × 5 = 2000 m

This is the correct answer but not a very neat method. This question was meant to be figured out using algebra. Please advise on an algebraic method. Thank you

 

[MathsFormula]

[1]@5.........d...........>t seconds

[2]<...........d............@4
900-t seconds

Hello Denis,
Unfortunately I don't understand what you are saying in the first two lines.

I think you're saying "at 5 m/s the distance travelled takes t seconds. Then the next line makes less sense to me.

Please guide me a little more. Thanks.

 

[Ishuda]

This is a maths algebra question and not physics so I don't think I'm meant to know the formula speed = distance / time so I won't use it.

Question says : A man runs to a
telephone and back in 900 seconds. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone.

My method :

900 seconds is divided in the ratio 5:4

= 500 : 400

So has been travelling 400 seconds to the telephone and 500 seconds on the way back.

Therefore distance to the telephone is 400 seconds travelled at 5 m/s = 400 × 5 = 2000 m

This is the correct answer but not a very neat method. This question was meant to be figured out using algebra. Please advise on an algebraic method. Thank you

Just as a matter of clearing something up for me: You did use speed=distance/time formula but in the rearranged form distance= speed*time. When asked 'math' questions such as this, you are many times expected to use some physics formulas.

Anyway, to get the time you can use the two formulas [akin to those of Denis and yours]
t₁ + t₂ = 900
t₁/t₂ = 5/4
That will give you the times envolved and you can use the distance formula to get the distance.

 

[MathsFormula]

Hello, sorry I couldn't get to replying sooner.

I've used bits of the ideas of from Den²is and Ishuda to get to my own solutions.

But I had to specifically use the physics formula speed = distance / time

t₁ = time to the telephone.

t₁ + t₂ = 900 ... (iii)


Speed = distance / time

5 = d/t₁ .... (i)

4 = d/t₂ .... (ii)

From (i) and (ii)

5t₁=4₂

Solving simultaneous equations (i) (ii) and (iii) t
we get t₁ = 400


Speed = distance / time

So 5 = distance / 400

Distance = 2000 m


I don't know how I would have done the question without knowing the formula. I can just about see a method but can'tcan't quite decipher it (it's at the tip of my tongue you could say).

I tried Ishuda method of ratios which I thought was brilliant UNTIL I came to the answer.

t₁/t₂ = 5/4

t₂ = 4/5 (t₁)

Using t₁ + t₂ = 900 ... (iii)

And solving gives t₁ = 500 seconds.

Then I solved further and got distance = 2500m

If the answer wasn't on the book to tell me I'm wrong I never would have realised that it's impossible for the journey to the phone take longer time than the way back.

Have I done the ratio the wrong way or something?
Is my reasoning lacking. I know this is an elementary question but I've made it massive in my mind.

 

[MathsFormula]

Accidentally posted again

 

[Dale10101]

Pretend

This is a maths algebra question and not physics so I don't think I'm meant to know the formula speed = distance / time so I won't use it.

Question says : A man runs to a
telephone and back in 900 seconds. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone.

My method :

900 seconds is divided in the ratio 5:4

= 500 : 400

So has been travelling 400 seconds to the telephone and 500 seconds on the way back.

Therefore distance to the telephone is 400 seconds travelled at 5 m/s = 400 × 5 = 2000 m

This is the correct answer but not a very neat method. This question was meant to be figured out using algebra. Please advise on an algebraic method. Thank you

Pretend you know the distance to the pole, call it D.

The time to get to the pole is then D/(5 m/s).

The time to return is D/(4 m/s).

We know that D/(5 m/s) + D/(4 m/s) = 900 s

Solving for D gives you 2000 m. I think no matter how you solve the problem you need to know the definition of speed in terms of the units that it is measured in. BTW I like your original solution, most efficient.