A snail starts on her trip from a pumpkin patch to a strawberry field. One hour and 36 minutes later, a turtle leaves the pumpkin patch for the strawberry field and arrives at the same time as the snail. The distance between the pumpkin patch and the strawberry field is 52 m. Find the speed of the snail if it moves 32 m/hr slower than the turtle.

You want the speed of a snail if the time it takes to travel 52 m is 1 hour and 36 minutes longer than the time it takes a turtle traveling 32 m/h faster.

Time

The relation between time, speed, and distance is ...

time = distance/speed

If s represents the speed of the snail, then s+32 is the speed of the turtle. The relation between their times is ...

[tex]\dfrac{52}{s}=\dfrac{52}{s+32}+1.6[/tex]

where 1.6 hours is 1 hour 36 minutes.

Solution

Multiplying the equation by s(s+32), we have ...

52(s +32) = 52s +1.6s(s +32)

52s +1664 = 52s +1.6s² +51.2s . . . . . eliminate parentheses

1.6s² +51.2s -1664 = 0 . . . . . . . . . . put in standard form

s² +32s -1040 = 0 . . . . . . . . . . . divide by 1.6

This quadratic can be solved in any number of ways.

(s +32s +256) -1040 -256 = 0 . . . . . complete the square

(s +16)² = 1296 . . . . . . . . . . . . . . add 1296

s +16 = 36 . . . . . . . . . . . . . . positive square root

s = 20 . . . . . . . . . . . . . . subtract 16

The speed of the snail is 20 m/h.