# math

posted by on .

In a motorcycle race, one lap of the course is 650 m. At the start of the race, John sets off 4 seconds after Tom does, but John drives his motorcycle 5m/s faster and finishes the lap 2.5 seconds sooner than Tom does.

What is the speed at which each of them is driving?

What is the time taken by each of them to cover the distance.

• math - ,

Tom's speed = X m/s.
John's speed = (X + 5) m/s.

Tom's time = Ys.
John's time=(Y - 2.5)S.

Eq1: X*Y = 650m.
Y = 650/X.

Eq2:(X + 5)(Y - 2.5) = 650m.

In Eq2, substitute 650/X for Y:
(X + 5)(650/X - 2.5) = 650,
650 - 2.5X + 3250/X -12.5 = 650,
650 - 2.5X + 3250/X - 12.5 -650 = 0,
-2.5X + 3250/X -12.5 = 0,
Multiply each side by -X:
2.5X^2 - 3250 + 12.5X = 0,
Divide each side by 2.5:
X^2 + 5X - 1300 = 0,
Solve for X using Quad.Formula and get:
X = 33.64; X = -38.64.
Select positive value of X:
X = 33.64m/s = Tom's speed.
X*Y = 650,
33.64Y = 650,
Y = 19.3s = Tom's time.

X + 5 = 33.64 + 5 = 38.64m/s = John's
speed.
John's time = Y - 2.5 = 19.3 - 2.5 -
16.8S.

X - 20 = 0,

X + 25 =0,
X = -25.
Choose the positive solution:
X = 20m/s = Tom's speed.

X + 5 = 20 + 5 = 25m/s = John's speed.

XY = 650,
20Y = 650,
Y = 650 / 20 = 32.5s. = Tom's time.

Y -6.5 = 32.5 - 6.5=26s. = John's time.

;

• math - ,

OOPS!

• math - ,

Alternate Approach:

Tom's speed = X m/s.
Tom's time = Ys.

John's speed = (X + 5)m/s.
John's time = Y - 2.5 - 4 = Y - 6.5.

Eq1: XY = 650m. Y = 650/X.
Eq2: (X + 5)(Y - 6.5) = 6.5m.

Substitute 650/X for Y in Eq2:

(X + 5)(650/X - 6.5) = 650,
650 - 6.5X + 3250/X - 32.5 = 650,
650 - 6.5X + 3250/X -32.5 -650 = 0,
-6.5X + 3250/X -32.5 = 0,
Multiply each side by -X:
6.5X^2 - 3250 + 32.5X = 0,
Divide each term by 6.5:
X^2 + 5X - 500 = 0,
(X - 20)(X + 25) = 0,

X - 29 = 0,
X = 20.

X + 25 = 0,
X = -25.

Select positive value of X:
Tom's speed = X = 20m/s.
XY = 650,
20^Y = 650,
Tom's time = Y = 650/20 = 32.5s.

John's speed = X + 5 = 20 + 5 = 25m/s.
John's time = 32.5 - 2.5 = 30s.