Question

The position co-ordinate of a particle that is confined to move along a straight line is given by \[\LARGE{\color{blue}{x=2t^3-24t+6.}}\] where 'x' is measured from a convenient origin & 't' is in seconds. Determine the distance travelled by the particle during the interval from
"t=1 sec. to t =4 sec".

The solution to it was given by @Vaidehi09

Answer 1

Its answer is 74 meters.

I have a solution for this: -



First differentiate the given exp to get v(t) = 6t^2 - 24
now integrate this graph twice - first from 1 to 2, then from 2 to 4.
answer to first will come out as -10 and the second will give 64.
distance is total area under the graph. so ignore the -ve sign and add the 2 values.
so 10+64 = 74m


The 54 units is the displacement while 74 units is the distance.

i.e.;
displacement = -10+64 = 54m
while
\[\color{red}{distance = 10+64 = 74m.}\]



But would anybody show me how to integrate ?

Answer 2

Sorry I put this question in open-study feedback by mistake.

Answer 3

\[\int\limits_{1}^{2} 6t ^{2} - 24\] = [6t^3/3 - 24t] from 1 to 2
= [2t^3 - 24t] from 1 to 2
= 16 - 48 - 2 + 24
= -10

Answer 4

formula used is \[x ^{n} = x ^{(n+1)} / (n+1)\]

Answer 5

proceed the same way for the second integration as well.

Answer 6

u got it?

Answer 7

wouldn't we add 6 to first integration as a constant?

Answer 8

i gt it:)

Answer 9

bt plz see my doubt:)

Answer 10

oh ok ok
i gt it
i think it doesn't matter if we add any constant
becoz the distance would come only & only 74 meters.

Am i right @Vaidehi09 ???????

Answer 11

i didn't understand ur doubt...bt here's what happens:
\[\int\limits_{1}^{2} 6t ^{2} =6 \int\limits_{1}^{2} t ^{2} = 6 [t ^{3}/3] from 1 \to 2\]

Answer 12

what do u mean by 'add constant'...add where?

Answer 13

no; actually i wanted to ask that if we differentiate \[x=2t^3-24t+6.\] we get \[v=6t^2-24.\]but if we integrate \[v=6t^2-24.\]we get \[x=2t^3-24t+6.\]then why have u written only \[x=2t^3-24t.\]
why didn't u add 6?

Answer 14

oh.. that is because we are dealing with definite integral here. in definite int, we don't have the constant C like in indefinite int.

Answer 15

because in def int, the constant anyway cancels out when we substitute the upper and lower limit values. so yes, you're ryt...adding or not adding the constant won't affect our final answer here.

Answer 16

ok thanx a lot once again:)
I gt my answer with my full satisfaction now.

Answer 17

gr8!

Answer 18

^_^