A point moves in x-y plane according to vec r=3cos(4t)veci +3(1-sin4t) vecj. What is the distance travelled by the particle in 0.5 sec ?

2 Answers

5.048\ \text{unit}

Explanation:

The position of particle at any time t is given as

\vec r=3\cos(4t)\vec i+3(1-\sin(4t))\vec j

Now, the position of particle at time t=0 is given as

\vec r_1=3\cos(0)\vec i+3(1-\sin(0))\vec j=3\veci+3\vecj

The position of particle at time t=0.5 is given as

\vec r_2=3\cos(2)\vec i+3(1-\sin(2))\vec j=-1.248\veci+0.272\vecj

Now, the distance traveled by the particle in t=0.5 sec is given as

|\vecr_1-\vec r_2|

=|3\veci+3\vecj-(-1.248\veci+0.272\vecj)|

=|4.248\veci+2.728\vecj|

=\sqrt{4.248^2+2.728^2}

=5.048\ \text{unit}

Jul 6, 2018

sqrt3 " units"

Explanation:

Vector quantity position/displacement is:

vec r=3cos(4t)veci +3(1-sin(4t)) vecj

Differentiate to obtain vector quantity velocity:

d/(dt)vecr=vec v=-12sin(4t)veci -12cos(4t) vecj

= - 12(sin(4t)veci + cos(4t) vecj)

Scalar quantity speed dot s is:

dot s = abs(vecv) = sqrt(vecv*vecv) = sqrt12 = 2sqrt3

In any time period of 0.5" sec", the particle will move a distance :

s = 1/2 * 2sqrt3 = sqrt3 " units"