dave13 wrote:

niks18 wrote:

dave13 wrote:

Hi niks18

thanks for your comments, why so? i used 4 variables in two "t"s and two "x"s i just didnt denote x and t as x1 and x2 - but it can clearly be seen that 40t and 60t is the same as 40t1 and 60t2 Hi

dave13what you are assuming here is t1=t2=t. is this really the case?

do you know the values of t1, t2 or t. no

as per your equation 40t1+60t2/x1/40+x2/60, if you solve this you should get

(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2)

how are you going to proceed further from here?

Hi

niks18, and

Bunuel(40t1+60t2)/[(3x1+4x2)/120]=120*20(2t1+3t2)/(3x1+4x2) a minor correcton

I solved denominator seperatly but the point is that i didnt indicate x as x1 and x2 and t as t1 and t2 - was it a must ?

so i did this 3x+2x/120 = 5x/120 is total total time

than numerotor calculated 40t+60t = 100t is a total distance

average speed = total distance / total time

100t / (5x/120) = 100t/1 *120/5x = 12000t/5x this is how i solved it

you are saying i need additional equation for linking variables, but i seperatly expressed time and distance for one part and seperatly expressed time and distance for second part

so i i linked all these variables through formula of finding average speed = total distance divided by total time.

if so which additional equation am i missing and why ?

many thanks!

hi

dave13what you are doing is really confusing

. Nonetheless i'll take a final shot in explaining the calculation to you using the variables you used.

distance traveled in the first part \(=x_1\)

so time taken to travel first part \(t_1=\frac{x_1}{40}\)

distance traveled in the second part \(=x_2\)

so time taken to travel second part \(t_2=\frac{x_2}{60}\)

Hence total time taken to complete the journey \(= t_1+t_2=\frac{x_1}{40}+\frac{x_2}{60}=\frac{3x_1+2x_2}{120}\)------------(1)

Now let the total distance traveled be \(d\).

given distance traveled in first part \(x_1= x\)% of \(d=\frac{xd}{100}\)

so distance traveled during the second part \(x_2= d-\frac{xd}{100}=\frac{d(100-x)}{100}\)

Now substitute the value of \(x_1\) & \(x_2\) in equation (1) to get total time \(= \frac{\frac{3xd}{100}+\frac{2d(100-x)}{100}}{120}\)

=> total time \(= \frac{d(x+200)}{12000}\)

therefore average speed \(= \frac{total distance}{total time}\) \(=\frac{d}{\frac{d(x+200)}{12000}}=\frac{12000}{(x+200)}\)