help me pleaseeee

Question

Jennifer drove 525 miles to see her friend. On the return trip, she drove 8 mph faster, so she made the return trip in 2 hour less time. What was her speed on the way to friend's home

Answer 1

525=rt

525=(r+8)(t-2)

r=525/t

525=([525/t]+8)(t-2)

525=([525+8t]/t)(t-2)

525t=(525+8t)(t-2)

525t=525t+8t^2-1050-16t

0=8t^2-16t-1050

4t^2-8t-525=0

4*525=2*2*5*5*3*7=(2*5*5)*(2*3*7)=50*42 and 50-42=8

(2t-25)(2t+21)=0

2t-25=0

2t=25

t=25/2

t=12.5 hours

525=r*12.5

r=525/12.5

r=42 mph (answer)

Answer 2

Hi Bob,

 

The basic equation to use in this problem is

 

x = vt

 

where x is the distance traveled, v is the speed and t is the time.

 

Let's denote variables for the outbound trip with a subscript 1, so that

 

x₁ = v₁t₁

 

and let us denote the variables for the return trip with a subscript 2, so that the corresponding equation is

 

x₂ = v₂t₂

 

But, the outbound distance and the return distance are the same, or 525 miles. So, we can call that distance d, so that

x₁ = x₂ = d.

 

So, the equations become

 

d = v₁t₁

d = v₂t₂

 

We can eliminate d by setting the equations equal to each other, since they both equal d. So,

 

d = v₁t₁ = v₂t₂

 

We know in Jennifer's return trip, she rode 8 miles/hour faster and took 2 fewer hours. So, we can now link v₁ and v₂ as well as t₁ and t₂, as

 

v₂ = v₁ + 8, remembering that speeds are in miles per hour

t₂ = t₁ - 2, remembering that times are in hours

 

So, we can make substitutions in our combined equation:

 

v₁t₁ = v₂t₂

 

or

 

v₁t₁ = (v₁ + 8)(t₁ - 2)

 

We can multiply this out to give

 

v₁t₁ = v₁t₁ - 2v₁ + 8t₁ - 16

 

We can cancel out the v₁t₁ on each side of this last equation and then get

 

8t₁ - 2v₁ = 16

 

We can divide through by 2 to give

 

4t₁ - v₁ = 8

 

This is one equation and two unknowns. But, remember we had the original equation in v₁ and t₁, d = v₁t₁.

We now have two equations and two unknowns. We can eliminate one of the variables. Let's eliminate t₁. So, we do this by solving both of these equations for t₁ and then combining them and thus eliminating t₁.

 

We get

 

t₁ = (1/4)(v₁ + 8)

 

t₁ = d/v₁

 

So, we have

 

(1/4)(v₁ + 8) = d/v₁

 

or, multiplying through this equation by 4 we have

 

v₁ + 8 = 4d/v₁

 

We can now multiply through v1 to give a quadratic  equation in v₁.

 

(v₁)² + 8v₁ - 4d = 0

 

But, d = 525, so we have

 

(v₁)² + 8v₁ - 2100 = 0

 

This factors as

 

(v₁ + 50)(v₁ - 42) = 0

 

So, v1 = -50 mph, v1 = 42 mph

 

I imagined Jennifer originally riding to the right, where the right is positive. The negative sign would refer to her return velocity, which happens to be 8 mph larger than her outbound speed. So, her speed to her friend's house is 42 mph.

 

 

Answer 3

I will set this up for you and you can complete the solution.  Note that the distance travelled is the same both ways.

 

v = speed going to her friend

t = time to get to her friend

v + 8 = speed coming back

t - 2 = time to get back

 

d = vt = (v + 8)(t - 2) = 525

 

(i)  vt = 525

(ii) (v + 8)(t - 2) = 525

 

From (i) get t = 525/v and substitute into (ii) to get v.

 

(v + 8)(525/v - 2) = 525

 

Solve this for v.  The result will be in mph.  You can also get t from (i) by substituting your answer for v.  Can you take it from here?  Feel free to post a comment if you need more help.