The Student Room Group

A level maths (differential equations)

I was wondering if they made a mistake in the following question, any help will be much appreciated :smile:
(edited 3 years ago)
Reply 1
For part (i) I get

dv/dt = 2/(t+2)

for part 2, we integrate that to get:

v = 2ln(t +2) + C

when t = 6,

v = 6ln2 + C

have they made a mistake in ignoring C? we can work out C since we know when t = 0, v = 0 as the car starts from rest. which gives C to be -2ln2
Reply 2
Original post by aramis8
I was wondering if they made a mistake in the following question, any help will be much appreciated :smile:

I get a different answer but

Original post by aramis8
For part (i) I get

dv/dt = 2/(t+2)

for part 2, we integrate that to get:

v = 2ln(t +2) + C

when t = 6,

v = 6ln2 + C

have they made a mistake in ignoring C? we can work out C since we know when t = 0, v = 0 as the car starts from rest. which gives C to be -2ln2

I get 4ln2 but I've done it very quickly so could be wrong :smile:
Reply 3
Original post by davros
I get a different answer but


I get 4ln2 but I've done it very quickly so could be wrong :smile:

Thanks for your reply

I get the same, 6ln2 - 2ln2

am I correct in assuming v = 0 when t = 0?
Reply 4
Original post by aramis8
Thanks for your reply

I get the same, 6ln2 - 2ln2

am I correct in assuming v = 0 when t = 0?

we're told the particle starts from rest, so v = 0 when t = 0 :smile:
Reply 5
Original post by davros
we're told the particle starts from rest, so v = 0 when t = 0 :smile:

Thanks :smile:

Quick Reply

Latest