Question 4488
OK, I would first draw a distance-time graph for each journey, to help visualise the question. I cannot show this easily here though, so i shall just have to define some things:

First journey:
Let d = distance travelled before the break
Let {{{t[1]}}} = time taken to travel this distance
Let v = speed of this part of the journey
Let 30-d = distance travelled after the break
Let {{{5-(t[1]+0.5)}}} be the time taken to travel this second part
Let v/2 = speed of this final part of the journey


Second Journey
Let d+10 = distance travelled before the break
Let {{{t[2]}}} = time taken to travel this distance
Let v = speed of this part of the journey
Let 30-(d+10) = distance travelled after the break
Let {{{4-(t[2]+0.5)}}} be the time taken to travel this second part
Let v/2 = speed of this final part of the journey


You need to understand these definitions in the context of the question. If you do not, you will just see it as "magic" on my part and you will not learn from it.


The question wants v and d, so we will be looking to remove the 2 times, {{{t[1]}}} and {{{t[2]}}}.


Basically, all I shall use now is the definition of speed=distance/time. So, for the 2 parts of the first journey we get:


{{{v=d/t[1]}}} --eqn1 and {{{(v/2) = ((30-d)/(5-(t[1]+0.5)))}}} -- eqn2


Eqn1 can be re-written as {{{t[1]=d/v}}}


Now, eqn2 is worked on as follows (remember we want {{{t[1]}}}= so we can equate both versions together). So...


{{{v = (2(30-d)/(4.5-(t[1])))}}}
{{{(4.5-(t[1])) = (2(30-d)/v)}}}
{{{(t[1]) = 4.5 - (2(30-d)/v)}}}


So, we now know 2 different equations that both equal {{{t[1]}}, so these must be the same. Hence,


{{{(d/v) = 4.5 - (2(30-d)/v)}}}
{{{(d/v) = 4.5 - (60-2d)/v}}}
{{{(d/v) = 4.5 - 60/v + 2d/v}}}
{{{-d/v = 4.5 - 60/v}}}
{{{-d/v + 60/v = 4.5}}} --eqn3


Now we do the same for the second journey:


{{{v = (d+10)/(t[2])}}} and {{{(v/2) = (30-(d+10))/(4-(t[2]+0.5))}}}


Re-arrange both of these to get t2= then equate these and simplify to give (60/v) - (d/v) = 3.5 + (10/v) --eqn4


Now, you can see eqn3 and eqn4 have the same lefthand side, so this means that their righthand sides are the same, ie 4.5 = 3.5 + 10/v


so, 1 = 10/v --> v=10


Now, put this value back into eqn3 or eqn4 to find d.


Hope this all makes sense.


jon.
PS there is an error on this page - i cannot seem to fix it. Email Igor to ask him to fix it, then the page will view correctly.