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

When Michael travels with slower speed his time is t1=9x.t_1=\dfrac{9}{x}.

When Michael travels with faster speed his time is t2=9x+3.t_2=\dfrac{9}{x+3}.

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


t1t2=14t_1-t_2=\dfrac{1}{4}

9x9x+3=14\dfrac{9}{x}-\dfrac{9}{x+3}=\dfrac{1}{4}

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

x2+3x108=0,x>0x^2+3x-108=0, x>0

D=(3)24(1)(108)=441D=(3)^2-4(1)(108)=441

x=3±4412(1)=3±212x=\dfrac{-3\pm\sqrt{441}}{2(1)}=\dfrac{-3\pm21}{2}

Since x>0,x>0, we take


x=3+212x=\dfrac{-3+21}{2}

x=9x=9

Therefore Michael's faster speed is 1212 km/h.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS