Question

a particle moves according to the equation x=acosPi t the distance covered by it in 2.5 sec is

Answer 1

x=a*cos(pi*t)
x=a*cos(2.5*pi)=?

Answer 2

answer is 5a -_-

Answer 3

x=a*cos(2*pi+0.5pi)=a*cos0.5*pi=a*0=?

Answer 4

Using the arclength formula and recognizing that \(\sin(\pi x)<0\) when \(1<x<2\) and positive otherwise in the interval \([0,2.5]\)
$$
x(t)=a\cos\pi t\\
x'(t)={d\over dt}x(t)=-a\pi \sin\pi t\\
|x'(t)|=a\pi\sin \pi t \\
\text{Which means, }\\
a\pi\sin(x)=\left\{\begin{matrix}
a\pi\sin(x)&0<x<1\\
-a\pi\sin(x)&1<x<2\\
a\pi\sin(x)&2<x<2.5\\
\end{matrix}\right.\\
\text{So, }\\
\int_0^{2.5}\left |x'(t)\right | dt=\int_0^{2.5}\left |a\pi \sin \pi t\right | dt\\
=a\pi\left ( \int_0^{1} \sin \pi t dt-\int_1^{2} \sin \pi t dt+\int_2^{2.5} \sin \pi t dt\right )\\
=a\pi\left (\left .{-cos\pi t\over \pi}\right |_0^1+\left .{cos\pi t\over \pi}\right |_1^2+\left .{-cos\pi t\over \pi}\right |_2^{2.5}\right)\\
=a\left ((-(-1)-(-1))+(1-(-1)-(0-1)\right)\\
=5a
$$

Answer 5

Here is an interactive chart of your problem I've created for you -

http://tube.geogebra.org/student/m843545

Let me know if you have any questions.

Answer 6

(Note that there may be a delay in the calculation of Arclength in the link above because the calculation is actually occurring out in a "cloud" server" - just wait a few seconds for the results)

Answer 7

(Note that there may be a delay in the calculation of Arclength in the link above because the calculation is actually occurring out in a "cloud" server" - just wait a few seconds for the results)

Answer 8

@ybarrap your math is actually calculating the area under the curve, not an **arc length**. in a proper arc length integral you would not need to split out the integration intervals in order to make the number work. in an arc length integration ds = sqrt ( x' ^ 2 + y' 2).

@kittycat01 if you start with the thing this question really deals with, namely simple harmonic motion, you should see the answer below.

i will rip that picture again in the next post and highlight the distance thing as that seems to be the only way to show different colours..