Answer to Question #236101 in Linear Algebra for quack

Question #236101

Michael is a bicycle commuter. he has observed that the reduces the time for his 9 km commute by 15 min when he increases his average speed by 3 km/h. what is Michael's faster speed?


1
Expert's answer
2021-09-12T23:52:28-0400

Let "x=" Michael's slower speed. Then his faster speed will be "x+3."

When Michael travels with slower speed his time is "t_1=\\dfrac{9}{x}."

When Michael travels with faster speed his time is "t_2=\\dfrac{9}{x+3}."

Michael reduces the time for his 9 km commute by 15 min


"t_1-t_2=\\dfrac{1}{4}"

"\\dfrac{9}{x}-\\dfrac{9}{x+3}=\\dfrac{1}{4}"

"36(x+3)-36x=x(x+3)"

"x^2+3x-108=0, x>0"

"D=(3)^2-4(1)(108)=441"

"x=\\dfrac{-3\\pm\\sqrt{441}}{2(1)}=\\dfrac{-3\\pm21}{2}"

Since "x>0," we take


"x=\\dfrac{-3+21}{2}"

"x=9"

Therefore Michael's faster speed is "12" km/h.



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