Math Class on February 23, 2010

In class today we worked on problem 3.2 on page 35 in groups. It shows at the top of the page how far Sascha traveled by bike after certain amounts of time. He stopped 3 times to record his time and distance. By stop 1 he had gone 5 miles in 20 minutes. 5/20 =15/60, which means 15 miles in 60 minutes, or 1 hour. This means he went 15 MPH for the first portion of the ride. By the second stop he had went 8 miles in 24 minutes. 8/24= 1/3= 20/60. That means 20 miles in 60 minutes, or 20 MPH. For the last one he went 15 miles in 40 minutes. 15/40= 22.5/60. That means 22.5 miles in 60 minutes, or 22.5 MPH. It is also okay to put the minutes as the numerator and the miles as the denominator. As you can tell, Sascha traveled the fastest at the third stop, and slowest at the first stop. If you were going 13 MPH on a bike it would take you 2 hours and 9 minutes to travel as far as Sascha did. Sascha traveled 28 miles (15+5+8). Our group decided to figure out how long it would take us to travel 1 mile going 13 MPH. 60/13=4.6. Since this is how long it would take us to travel 1 mile, we then multiply 4.6 by 28. 4.6*28=129.23. Then we divided that by 60 and got 2.15. The .15 represents 15/100 which equals 6/60, or 2 hours and 9 minutes. To find Sascha's steady rate, you find his average speed by adding 15+20+ 22.5=57.5. Divide this by 3 to find the average speed.