An employee drove to work on Monday at 45 mi/h and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mi/h and arriving one minute late.
a. How far does the employee live from work?
b. At what speed should the employee drive to arrive five minutes early?
Using V = D/T
60D/45 = T - 1
60D/40 = T + 1
Solving for T and equating yields
60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles
In order to arrive in 17 minutes,
60(12)/V = 17 yielding = 42.35mph.
To find the distance from the employee's home to work, let's calculate the time it takes for the employee to drive on Monday and Tuesday.
a. Distance from home to work:
Let's assume the distance from home to work is "d" miles.
On Monday:
The speed is 45 mi/h, and the employee arrived 1 minute early.
Time taken on Monday = d / 45 hours
Convert 1 minute to hours: 1 minute / 60 minutes = 1/60 hours
Time taken on Monday = d / 45 - 1/60 hours
On Tuesday:
The speed is 40 mi/h, and the employee arrived 1 minute late.
Time taken on Tuesday = d / 40 hours
Convert 1 minute to hours: 1 minute / 60 minutes = 1/60 hours
Time taken on Tuesday = d / 40 + 1/60 hours
Since the employee travels the same distance, we can equate the time taken on Monday and Tuesday:
d / 45 - 1/60 = d / 40 + 1/60
To solve this equation, let's first find a common denominator:
Multiply both sides by 45 * 40 * 60 to eliminate the denominators:
(40 * 60)d - (45 * 60 * d) = (45 * 40 * 60 * d) + (40 * 60)
(2400)d - (2700)d = (2700)(2400)d + (2400)
-300d = 6480000d + 2400
6480300d = -2400
Divide both sides by -300:
d = -2400 / -300
d = 8
Therefore, the employee lives 8 miles away from work.
b. To calculate the speed at which the employee should drive to arrive 5 minutes early, we can use the formula:
Time = Distance / Speed
Given that the employee needs to arrive 5 minutes early, we have:
Time = (d / Speed) - 5 minutes
Let's substitute d = 8 and solve for Speed:
(d / Speed) - 5 = 0
8 / Speed - 5 = 0
8 = 5 * Speed
Speed = 8 / 5
Therefore, the employee should drive at a speed of 8/5 mi/h (1.6 mi/h) to arrive 5 minutes early.