distant rate time word problem

Question

Scott took a trio of 226 miles. He averaged 54 miles per hour for the first part of the trip and 58 miles per hour for the second part.if the trip took four hours, how long did he travel at each rate?

Answer 1

Let's say he drove for t1 hours at 54 miles per hour, and t2 hours at 58 miles per hour.  Since he drove for a total of 4 hours, we have

 

t1 + t2 = 4.

 

Next, he drove for 226 miles.  Let's start with a definition of speed as:

(speed) = (distance)/(time).  Multiplying both sides by time, we have

 

(distance) = (speed)x(time).

 

Since he drove for t1 hours at 54 miles per hour, the distance he covered during that time (which we'll call d1) is

d1 = (54)t1.

 

Likewise, since he drove for t2 hours at 58 per hour, the distance he covered during that time (d2) is

d2 = (58)t2

 

If we add the distances, we get the total distance of the trip, or 226 miles:

d1 + d2 = 226

 

Substitute in our equations for d1 and d2:

 

(54)t1 + (58)t2 = 226

 

this produces a system of equations, one from the total time as being the sum of t1 and t2, and the other from the total distance being the sum of d1 and d2:

 

t1 + t2 = 4

(54)t1 + (58)t2 = 226.

 

Solve by substitution (using the first equation) or by elimination.

Answer 2

x = time at 54 mph

y = time at 58 mph

distance = rate*time

 

x + y = 4               equation 1

226 = 54x + 58y    equation 2

 

Using substitution from equation 1, y = 4-x

Substitute that into equation 2 to solve for x

 

226 = 54x + 58(4-x)

226 = 54x + 232 - 58x

226 = -4x + 232

subtract 232 from both sides

-6 = -4x

divide both sides by -4

-6/(-4) = x

3/2 = x

 

x + y = 4

y = 4-x

y = 4-3/2

y = 5/2 = 2 1/2

 

He drove 1 1/2 hours at 54 mph

and 2 1/2 hours at 58 mph