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 ?

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Answer 1 · 2018-07-05T22:35:19

#5.048\ \text{unit}#

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}#

Answer 2 · 2018-07-06T10:22:12

# sqrt3 " units"#

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"#