AS mechanics question (maths)

Hello, I have attempted this question however my method leads to acceleration giving imaginary solutions... I tried finding the magnitude of acceleration by doing Pythagoras (e.g. square root of (6t-15)^2 +16.. What am I doing wrong?

Can you post your working?

I know this is probably very wrong but this is what I've tried to do

Use F = ma and we know m = 2kg and F has magnitude 10

that is what I was trying to do, hence why I was trying to find the magnitude of acceleration. However what I've done gives no real solutions, could you explain a little more in detail? thank you

So what is the accel equal to? You just sem to have an expression.

imaginary solutions

imaginary solutions

I think what muttley was getting at is that we need to see the next stage of your working i.e. what equation are you trying to solve?

Well I wasn't sure what next step to take since I though that by finding the value for a by solving the quadratic I could go on to substitute it into F=ma

Well I wasn't sure what next step to take since I though that by finding the value for a by solving the quadratic I could go on to substitute it into F=ma

Have you managed to solve this yet? Your quadratic in t is correct if you equate it correctly to (the value of) a^2 and solve as mentioned by Muttley and Davros. There is a nice shortcut as well, but its worth doing the question the "standard way" first.

no I'm struggling to solve it since the quadratic equation keeps giving me imaginary solutions for t. Can someone tell me what the quadratic equation is meant to be!

What did you get/use for a? f = 10N and m = 2kg. Its almost impossible not to get the right value for a?

wait I've worked it out now, I had a bit of a misunderstanding before! thank you! if that's alright could you expand on what you mean by a shortcut?

Sure, what did you get (and how)?

I did 25=36t^2-180t+241 and solved it as a quadratic

Sure, thats the obvious way.
Once you have a=5, you could note that you have a 3(2t-5) : 4 : 5 right angled triangle so the first side must be +/-3 so
2t-5 = +/-1
which gives the two solutions.
As a middle ground solution, don't expand and square the 3(2t-5) term, just apply pythagoras to get
9(2t-5)^2 + 16 = 25
and that gives 2t-5 = +/-1 again.
Sometimes its worth keeping things factorized.

thank you!