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?
speed problem
- Thread starter 12345678
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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!
His original speed is \(\displaystyle s\) km/hr.
In \(\displaystyle t\) hours, he travels 400 km: .\(\displaystyle st = 400 \quad\Rightarrow\quad t = \frac{400}{s}\) .[1]
If his speed is \(\displaystyle (s+2)\) km/hr, it will take him \(\displaystyle \left(t-\tfrac{1}{3}\right)\) hours.
. . \(\displaystyle (s+2)(t-\tfrac{1}{3}) \:=\:400 \quad\Rightarrow\quad st - \tfrac{1}{3}s + 2t - \tfrac{2}{3} \:=\:400\)
Multiply by 3: .\(\displaystyle 3st - s + 6t - 2 \:=\:1200 \quad\Rightarrow\quad 3st - s + 6t \:=\:1202\)
Substitute [1]: .\(\displaystyle 3s(\frac{400}{s}) - s + 6(\frac{400}{s}) \:=\:1202\)
. . . . . . . . . . . . . . .\(\displaystyle 1200 - s + \frac{2400}{s} \:=\:1202\)
. . . . . . . . . . . . . . . . . . .\(\displaystyle -s + \frac{2400}{s} \:=\:2\)
Multiply by \(\displaystyle s\!:\;\;-s^2 + 2400 \:=\:2s\)
. . . . . . . . \(\displaystyle s^2 + 2s - 2400 \:=\:0\)
. . . . . . \(\displaystyle (s-48)(s+50) \:=\:0\)
Hence: .\(\displaystyle s = 48,\;\;\color{red}{\rlap{///////}}s=-50\)
Therefore, his original speed is \(\displaystyle 48 \text{ km/hr.}\)
His original speed is \(\displaystyle s\) km/hr.
In \(\displaystyle t\) hours, he travels 400 km: .\(\displaystyle st = 400 \quad\Rightarrow\quad t = \frac{400}{s}\) .[1]
If his speed is \(\displaystyle (s+2)\) km/hr, it will take him \(\displaystyle \left(t-\tfrac{1}{3}\right)\) hours.
. . \(\displaystyle (s+2)(t-\tfrac{1}{3}) \:=\:400 \quad\Rightarrow\quad st - \tfrac{1}{3}s + 2t - \tfrac{2}{3} \:=\:400\)
Multiply by 3: .\(\displaystyle 3st - s + 6t - 2 \:=\:1200 \quad\Rightarrow\quad 3st - s + 6t \:=\:1202\)
Substitute [1]: .\(\displaystyle 3s(\frac{400}{s}) - s + 6(\frac{400}{s}) \:=\:1202\)
. . . . . . . . . . . . . . .\(\displaystyle 1200 - s + \frac{2400}{s} \:=\:1202\)
. . . . . . . . . . . . . . . . . . .\(\displaystyle -s + \frac{2400}{s} \:=\:2\)
Multiply by \(\displaystyle s\!:\;\;-s^2 + 2400 \:=\:2s\)
. . . . . . . . \(\displaystyle s^2 + 2s - 2400 \:=\:0\)
. . . . . . \(\displaystyle (s-48)(s+50) \:=\:0\)
Hence: .\(\displaystyle s = 48,\;\;\color{red}{\rlap{///////}}s=-50\)
Therefore, his original speed is \(\displaystyle 48 \text{ km/hr.}\)
cheers for your help