speed problem

12345678

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Another problem: "On a journey of 400km, the driver of the train calculates that if he were to increase his average speed by 2km/h, he would take 20 minutes less. Find his average speed". Speed=Distance/time. (speed=s, time=t). S+2=400km/t-20. I then got rid of the denominator: (s+2)(t-20)=400. Therefore st-20s+2t-40=400km. Then I got rid of -40, so st-20s+2t=440km. I then made it speed=, so s=440+20s-2t/t. I was wondering what I am suppose to do next?
 
Another problem: "On a journey of 400km, the driver of the train calculates that if he were to increase his average speed by 2km/h, he would take 20 minutes less. Find his average speed".

Speed=Distance/time. (speed=s, time=t). S+2=400km/(t-20). I then got rid of the denominator: (s+2)(t-20)=400. Therefore st-20s+2t-40=400km. Then I got rid of -40, so

st-20s+2t=440km. Now use s*t = 400 & t = 400/s


I then made it speed=, so s=440+20s-2t/t. I was wondering what I am suppose to do next?
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s*t=400km. So 440+20s-2t=400km. 40+20s-2t=0. If this is what you meant I am not sure what to do next, as I cannot factorise this as it is not in the form ax^2+bc+c. Also, T=400/s. so T=400/((440+20s-2t)/t). If I have done correctly what you said, I still don't understand how to find his average speed?
 
Hello, 12345678!

On a journey of 400km, the driver of the train calculates that if he were to increase
his average speed by 2km/h, he would take 20 minutes less. .Find his original speed.

His original speed is s\displaystyle s km/hr.
In t\displaystyle t hours, he travels 400 km: .st=400t=400s\displaystyle st = 400 \quad\Rightarrow\quad t = \frac{400}{s} .[1]

If his speed is (s+2)\displaystyle (s+2) km/hr, it will take him (t13)\displaystyle \left(t-\tfrac{1}{3}\right) hours.

. . (s+2)(t13)=400st13s+2t23=400\displaystyle (s+2)(t-\tfrac{1}{3}) \:=\:400 \quad\Rightarrow\quad st - \tfrac{1}{3}s + 2t - \tfrac{2}{3} \:=\:400

Multiply by 3: .3sts+6t2=12003sts+6t=1202\displaystyle 3st - s + 6t - 2 \:=\:1200 \quad\Rightarrow\quad 3st - s + 6t \:=\:1202


Substitute [1]: .3s(400s)s+6(400s)=1202\displaystyle 3s(\frac{400}{s}) - s + 6(\frac{400}{s}) \:=\:1202

. . . . . . . . . . . . . . .1200s+2400s=1202\displaystyle 1200 - s + \frac{2400}{s} \:=\:1202

. . . . . . . . . . . . . . . . . . .s+2400s=2\displaystyle -s + \frac{2400}{s} \:=\:2


Multiply by s ⁣:    s2+2400=2s\displaystyle s\!:\;\;-s^2 + 2400 \:=\:2s

. . . . . . . . s2+2s2400=0\displaystyle s^2 + 2s - 2400 \:=\:0

. . . . . . (s48)(s+50)=0\displaystyle (s-48)(s+50) \:=\:0

Hence: .s=48,    ///////s=50\displaystyle s = 48,\;\;\color{red}{\rlap{///////}}s=-50


Therefore, his original speed is 48 km/hr.\displaystyle 48 \text{ km/hr.}
 
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