Intermediate AlgebraHeavy traffic reduced a driver’s usual speed by 10 mph, which lengthened his 200-mile trip by 1 hour. Find the driver’s usual speed.

Question

1. Heavy traffic reduced a driver’s usual speed by 10 mph, which lengthened his 200-mile trip by 1 hour. Find the driver’s usual speed. 

Answer 1

My approach would be to define terms up front:

D = distance

S_u = usual speed

T_u = usual time

S_t = today's speed

T_t = today's time

Then write out any relationships you know. For example, for either day, the distance will equal the product of speed and time, so we can write the following two equations:

D = S_u * T_u

D = S_t * T_t

We also know:

D = 200

Placing D on the right-hand side, and substituting the value of 200, our equations become:

Equation 1: S_u * T_u = 200

Equation 2: S_t * T_t = 200

That's two equations in four unknowns, which is not solvable. So we need more information. Fortunately, we also know that today's speed is 10 mph slower than usual:

S_t = S_u - 10

And we know that today's time is 1 hour more than usual:

T_t = T_u + 1

Now we need to put this all together to get the desired answer, namely the usual speed, S_u. Let's take our two equations in four unknowns and manipulate them into two equations in just two unknowns. We can do this by substituting these additional pieces of information into Equation 2 to get a new version, Equation 2B:

Equation 2B: (S_u - 10) * (T_u + 1) = 200

Multiplying this out and combining constant terms we get

S_u * T_u - 10T_u + S_u - 10 = 200

S_u * T_u - 10T_u + S_u = 210

Now our original Equation 1 and our new Equation 2B form a system of two equations in two unknowns:

Equation 1: S_u * T_u = 200

Equation 2B: S_u * T_u - 10T_u + S_u = 210

We need another substitution to get rid of T_u in one of these equations. Let's manipulate Equation 1 to isolate T_u:

T_u = 200 / S_u

Now substitute this into Equation 2B wherever we see T_u:

S_u * (200 / S_u) - 10(200 / S_u) + S_u = 210

We now have one equation in one unknown. Getting very close! Simplify this a bit:

200 - (2000 / S_u) + S_u = 210

S_u - (2000 / S_u) - 10 = 0

To get S_u out of the denominator in (2000 / S_u), multiply both sides of the equation by S_u:

S_u² - 10S_u - 2000 = 0

This equation is now in fighting trim. Solving for S_u means cracking the quadratic polynomial, either using the quadratic equation, or by factoring. In this case, factoring is possible, giving:

(S_u + 40)(S_u - 50) = 0

Mathematically this has two "zeroes" or solutions;

S_u + 40 = 0 ==> S_u = -40

OR

S_u - 50 = 0 ==> S_u = 50

Since we know it is not possible for your usual speed to be negative 40 miles per hour (this is just speed, not directed velocity), we can treat 50 mph as the correct answer.

S_u = 50 mph

Let's check our answer out. Using S_u = 50, solve our original Equation 1 for T_u, our usual time:

T_u = (200 / S_u) = 200/50 = 4 hours

Now take 10 mph off the usual speed S_u to get today's speed S_t:

S_t = 40 mph

Based on this, today's time T_t time for today would be:

T_t = (200 / S_t) = 200/40 = 5 hours.

That checks out! So S_u, the usual speed, is verified to be 50 mph.

Answer 2

The difference in speed is 10 mph. we write the speed in both cases (usual or traffic).

V1 - V2 = 10

200 / t - 200 / (t+1) = 10, where t is the time required for 200 miles with usual speed. Now just solve this rational equation by taking the common denominator. First divide both sides by 10 to reduce the numbers:

20 / t - 20 / (t + 1) = 1

(20 t + 20 - 20 t) / (t (t + 1)) = 1 Cancel 20t

20/ (t(t + 1)) = 1 and then cross multiply:

20 = t ( t + 1)

t^2 + t - 1 = 0 Now factor the second degree equation

(t - 4)( t + 5) = 0 Then put each factor = 0

t - 4 = 0 So: t = 4 which is the hour at usual speed.. ( t = -5 is not acceptable since it is negative)

So the driver's usual speed is V = X / t = 200 / 4 = 50 mph

Answer 3

Usual speed was 50 mph, in 4 hours covering 200 miles

slower speed was 40mph in 5 hours covering 200 miles

UT = 200

(U-10)(T+1) = 200

UT -10T +U -10 =200 = UT

U-10T = 10

U -10(200/U) = 10

U^2 -10U = 2000

(U-5)^2 = 2025

U = 5+45 = 50 mph for usual speed