Jon left his house and headed toward Brandon's house at the same time Brandon headed toward Jon's house. Brandon traveled 50 mi/h and Jon traveled 55m/h. The distance between their homes is 70 miles. How long did they travel before they met?
Please help me wih this problem. Thank you.
50*t+55*t=70
solve for t.
this is what i got im not sure if its right.
50t+55t=70
50t+55t=105t=70
70/105
t=1.5
To solve this problem, you can use the concept of relative speed.
Let's break down the information given:
- Brandon is traveling at a speed of 50 miles per hour (mi/h).
- Jon is traveling at a speed of 55 miles per hour (mi/h).
- The distance between their homes is 70 miles.
Since they both started at the same time and are heading towards each other, their combined speed will determine how quickly they meet.
To find the combined speed, add their individual speeds:
Combined speed = Brandon's speed + Jon's speed
Combined speed = 50 mi/h + 55 mi/h = 105 mi/h
Next, we can use the formula "distance = speed × time" to find the time it takes for them to meet.
Let's assume they meet after time t (in hours).
For Brandon:
Distance traveled by Brandon = Brandon's speed × time = 50t miles
For Jon:
Distance traveled by Jon = Jon's speed × time = 55t miles
Since they meet when the sum of their distances is the total distance between their homes (70 miles), we can set up the equation:
Distance traveled by Brandon + Distance traveled by Jon = Total distance
50t + 55t = 70
Combining like terms:
105t = 70
Divide both sides by 105:
t = 70 / 105
Simplifying:
t = 2/3
Therefore, they traveled for 2/3 of an hour, or approximately 40 minutes, before they met.