A teacher drove 280 miles to attend a mathematics conference and arrived 1 hour late. The teacher figured out that, had she increased her average speed by 5 mph, she would have arrived on time to the conference. What was her average rate of speed?
*explanations would be awesome if possible because i have a test on this tomorrow and have NO clue how to do it..thanks!
Let T be the time of the trip in hours, if one arrives on time. Let V be the speed he first time.
280/V = T+1
280/(V+5)= T
Solve the two equations for the two unknowns.
1/V - 1 = 1/(V+5)
1(V+5)/V - (V+5) = 1
1 + 5/V - V -5 = 1
-V^2 -5V + 5 = 0
Take the positive root of the quadratic equation.
Check my thinking
To find the teacher's average rate of speed, we need to set up a mathematical equation based on the information given.
Let's assume her original average speed is represented by x mph. Since she arrived 1 hour late, we can say that the time it took her to travel the distance of 280 miles at x mph is 280 / x hours.
If she had increased her average speed by 5 mph, her new average speed would be (x + 5) mph. At this speed, she would have arrived on time, so we can say that the time it would have taken her to travel the same distance at (x + 5) mph is also 280 / (x + 5) hours.
We are given that the difference in time is 1 hour. So we can set up the following equation:
280 / x - 280 / (x + 5) = 1
Now, let's solve the equation step by step:
1. Multiply both sides of the equation by x(x + 5) to eliminate the denominators:
280(x + 5) - 280x = x(x + 5)
2. Distribute the multiplication on both sides:
280x + 1400 - 280x = x^2 + 5x
3. Simplify and simplify further:
1400 = x^2 + 5x
4. Rearrange the equation to form a quadratic equation by bringing all the terms to one side:
x^2 + 5x - 1400 = 0
Now, we can either factor this quadratic equation or use the quadratic formula to find the values of x.
If we factor it, the equation would become:
(x + 40)(x - 35) = 0
Setting each factor to zero, we find two potential solutions:
x + 40 = 0 or x - 35 = 0
Solving each equation, we find that:
x = -40 or x = 35
Since the average speed cannot be negative, we discard x = -40. Therefore, the teacher's average rate of speed is 35 mph.