Dan rides his bicycle 15 miles at one speed. He reduces his speed by 2 mph and continues on for 8 miles more. If his total time traveled is 2-1/2 hours...

Question

Dan rides his bicycle 15 miles at one speed. He reduces his speed by 2 mph and continues on for 8 miles more. If his total time traveled is 2-1/2 hours, what was his original speed?

 

PLEASE HELP AND SHOW ALL WORK

Answer 1

Danny,

The basic equation for this problem is d = vt, or distance = velocity * time. What they gave you were two different partial distances that add up to the total distance traveled, a difference between the two velocities involved, and the total time involved. There are multiple ways you could set up this problem, but I'll do the one I think will be easiest to follow.

 

d₁ = distance of the first leg

d₂ = distance of the second leg

v₁ = velocity of the first leg

v₂ = velocity of the second leg

t₁ = time of the first leg

t₂ = time of the second leg

 

d₁ + d₂ = v₁t₁ + v₂t₂

 

15 + 8 = v₁t₁ + v₂t₂    You can see that there are 4 different variables to solve for, which is impossible with one equation. But the additional information we have does relate v₁ to v₂ and t₁ to t₂. With that information, we can get it down to two variables. But we still need two equations to solve those. So rather than lump both legs into a single equation, we'll write each leg as a separate equation and solve the system.

 

Relationships given:   v₂ = v₁ - 2 ,       t₁ + t₂ = 2.5  ∴    t₂ = 2.5 - t₁

 

First leg:       15 = v₁t₁    ∴     t₁ = 15/v₁

Second leg:    8 = v₂t₂ = (v₁ - 2)(2.5 - t₁)

 

You can see we have two equations using v₁ and t₁. So we can solve the system.

First, expand the second leg:   8 = 2.5v₁ - v₁t₁ - 5 + 2t₁

Knowing that the question has asked for his original speed, t₁, we want to solve for that first to save time.

We can substitute into the second leg equation from the first leg equation in a way that leaves us with v₁.

 

8 = 2.5v₁ - 15 - 5 + 2(15/v₁)         Now combine like terms

 

0 = 2.5v₁ + 30/v₁ - 28                  Multiply the equation by v₁ to get rid of the denominator.

 

0 = 2.5v₁² - 28v₁ + 30                  Use the quadratic formula to solve.

 

v₁ = [28 ± √(28² - 4(2.5)(30))] / 2(2.5)

 

v₁ = (28 ± 22)/5

 

v₁ = 10  or   6/5      Now you might think, because they are both positive, that both are valid answers. But in fact, if you use v₁ = 6/5, then v₂ would be -4/5, and a negative velocity isn't really appropriate for this scenario. So your answer is:

 

v₁ = 10 mph

 

I hope that helps.