CCSS.MATH.CONTENT.HSA.CED.A.1
Create equations and inequalities in one variable and use them to solve problems.Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

Problem:

A mountain climber is scaling a 300-foot rock face at a constant rate. The climber starts at the bottom at 12:00 P.M. and by 12:30 P.M., the climber has moved 62 feet up the wall.

1. Write an equation that gives the distance d (in feet) remaining in the climb in terms of t (in hours).
2. At what time will the mountain climber reach the top of the rock face?

 

Commentary:

This problem deals with solving linear equations. The main thing is that the problem requires students to write their own equation. Students will have to interpret the word problem and write their own equation. This is good because it helps students better understand what they are solving for. This problem is an effective problem because it incorporates both writing the equation and then solving it at a certain value, in this case the top of the mountain. This problem allows students to identify the information given to make an equation and it requires them to use that equation to solve the problem.

Solution:

A)     d= distance (feet)

t= time (hours)

The climber climbs 62 feet in half an hour (12:00 P.M. to 12:30 P.M.) Since t is in hours then you need to double both quantities.

62*2=124

So the climber climbs 124 feet per hour.

To put this in an equation you need the distance (d) to equal the constant rate per hour (124) times t.

d=124t

 

B)      The rock climber will be at the top of the rock face when he is 300 feet high. So this means that d equals 300.

300=124t

Divide both sides by 124

So t equals 2.4 hours or 2 hours and 24 minutes