Question

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.

Answer 1

(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

Answer 2

s = 45 or -35
well, not going backwards so 45 miles/hour

Answer 3

https://www.mathsisfun.com/quadratic-equation-solver.html

Answer 4

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.

Answer 5

(V-10)(T+2)
slower takes longer :)

Answer 6

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

Answer 7

We agree :)

Answer 8

Yes! Thanks!