A bus driver, because of snow, was forced to drive at 10 miles an hour less than his usual speed, and he arrived 2 hours late at his destination 315 miles away. Find his usual speed.
(s-10)(t+2) = 315
s t = 315 so t = 315/s
(s-10)(315/s + 2) = 315
315 +2s - 3150/s - 20 = 315
2 s - 20 - 3150/s = 0
2 s^2 -20 s -3150 = 0
solve quadratic
a = 2 , b = -20 c = -3150
s = 45 or -35
well, not going backwards so 45 miles/hour
https://www.mathsisfun.com/quadratic-equation-solver.html
Usual speed = V mi/h.
Actual speed = (V-10) mi/h.
Eq1: V*T = 315.
(V-10)(T-2) = 315.
Since both Equations are = 315, they are also equal to each other:
V*T = (V-10)(T-2).
V*T = VT-2V-10T+20,
2V + 10T = 20,
V + 5T = 10,
V = -5T + 10.
In Eq1, replace V with -5T+10.
(-5T+10)T = 315,
-5T^2 + 10T - 315 = 0,
T^2 -2T - 63 = 0,
Solve by factoring or Quad. Formula.
Plug the value of T into Eq1 and solve for V.
(V-10)(T+2)
slower takes longer :)
Correction:Eq2: (V-10)(T+2) = 315.
V*T = (V-10)(T+2).
VT = VT+2V-10T-20,
-2V+10T = -20,
V - 5T = 10,
V = 5T + 10.
In Eq1, replace V with 5T+10:
(5T+10)T = 315,
5T^2 + 10T - 315 = 0,
T^2 + 2T - 63 = 0,
(T-7)(T+9) = 0,
T = 7, -9.
T = 7h.
V*T = 315.
V*7 = 315,
V = 45 mi/h.
=10
t
We agree :)
Yes! Thanks!
To solve this problem, we can use the formula: Distance = Rate × Time.
Let's assume the bus driver's usual speed is represented by "x" miles per hour.
Since the bus driver was forced to drive at 10 miles per hour less than his usual speed, his speed during this journey was (x - 10) miles per hour.
The distance traveled is given as 315 miles, and the time taken for the journey is 2 hours more than it should have been.
Using the formula, we can set up the equation:
Distance = Rate × Time
315 = (x - 10) × (Time + 2)
Now, let's simplify the equation:
315 = (x - 10) × (Time + 2)
315 = x × (Time + 2) - 10 × (Time + 2)
315 = x × Time + 2x - 10 × Time - 20
Simplifying further:
315 = x × Time - 10 × Time + 2x - 20
315 = (x - 10) × Time + 2x - 20
335 = (x - 10) × Time + 2x
We know that Time = 2, so substituting that in the equation:
335 = (x - 10) × 2 + 2x
335 = 2x - 20 + 2x
335 = 4x - 20
355 = 4x
x = 355 / 4
x = 88.75
Therefore, the bus driver's usual speed is 88.75 miles per hour.