Question

A particle moves in a straight line so that, t seconds after passing through a fixd point O, its velocity, v m/s, is given by \[v=\frac{60}{(3t+4)^{2}}\]

(iii) Find an expression for the displacement of the particle from O, t seconds after it has passed through O.

I will post my working, hopefully you'll tell me where I went wrong. A particle moves in a straight line so that, t seconds after passing through a fixd point O, its velocity, v m/s, is given by \[v=\frac{60}{(3t+4)^{2}}\]

(iii) Find an expression for the displacement of the particle from O, t seconds after it has passed through O.

I will post my working, hopefully you'll tell me where I went wrong. @Mathematics

Answer 1

\[\int\frac{60}{(3t+4)^2}dx=60\cdot\int(3t+4)^{-2}\]

Answer 2

\[=60[\frac{(3t+4)^{-1}}{(-1)(3)}]\]

Answer 3

\[=60[\frac{1}{3t+4}\cdot\frac{1}{-3}]\]

Answer 4

haha forgot the +C.
\[=60[\frac{1}{-9t-12}+C]\]

Answer 5

\[=\frac{60}{-9t-12}+60C\]
which is displacement against time right?
\[s=\frac{60}{-9t-12}+60C\]

Answer 6

what's the answer in the book

Answer 7

When t=0, s=0 also.
\[0=\frac{60}{-9(0)-12}+60C\]
\[0=\frac{60}{-12}+60C\]
\[60C=\frac{60}{12}\]

Answer 8

The answer in the book is:
\[s=5-\frac{20}{3t+4}\]

Answer 9

don't multiply 60 by c... c1+c2+c3=c same as 60c will just equal another constant

Answer 10

so you should get just c=60/12

Answer 11

thats where the 5 comes then above in your integral if you pull out a 3 from your denominator you should be able to reduce 6 to -20

Answer 12

continuing my working...
\[C=\frac{1}{12}\]
So the equation is...
\[s=\frac{60}{-9t-12}+60(\frac{1}{12})\]
\[s=\frac{60}{-9t-12}+5\]

Answer 13

Yeah, multiplying the C wasn't necessary, but it doesn't give a wrong answer.

Answer 14

on the denominator pull out a -3 and see what you get =]

Answer 15

technically your answer is correct just not fully simplified

Answer 16

thank your book for not giving all the forms that it could be lol

Answer 17

\[s=-\frac{20}{3t+4}+5\]

Answer 18

BALLS
That is correct!