Quick question

[vea]

just wondering
how would you guys do this question?
Is it wrong to let
t^2 + 1 = 7t - 5
then
2t = t + 6
then solve from there

You're right, you have to find a value(s) of t that satisfy both equations. Remember to sub t back in though cos the question is asking for the point of collision!

anyone know what the deal is with the abs. value signs in the logs?
wtf is that shit?
sometimes they just make what's in there positive and other times they leave it and i can't work it out

Leave them there unless you can actually justify when you're removing them. Trial exam papers are usually just lazy in their solutions!

[HarveyD]

hmm when I do that I get the other t value (which doesnt collide)
as t = 1

the answers below has t = 8 :S

should i use that method if a similar question comes up tomorrow?

[Natters]

will marks be taken off if i use them where theyre not needed? ive always just used brackets lol

[Zebra]

is this an exam 1 question?

[vea]

Oh yup if you want to find where the paths will meet (but not necessarily collide), you can do what they did or you can equate the Cartesian equations.

[HarveyD]

ahh alright
thanks heaps
good luck tomorrow!

and yeah, this is an exam 1 question

[Dangster]

i thought it was c, because i got a=0 if F=7. please correct me if i'm wrong? :\