00:01
Hi, in this question it is given that a particle is moving in a circular path with an angular speed of 6 .41 radius per second and the center of this circle is at origin and initially at time equal to 0 the particle is at x equal to 5 .94 meter and the y coordinate is 0 so since the particle is moving along the circular path then equation of circle is given as x squared plus y squared equal to r squared here r is the radius of this circle so in polar form this x is equal to r cos omega t and the y coordinate is r sine omega t since it is given that at t equal to at t equal to 0 second this x is equal to 5 .94 so here we can write r c c0 equal to 5 .9 4 so here we can write r coss 0 equal to 5.
01:02
94 so from here we can write this r value is equal to 5 .94 meter so radius of circle is 5 .94.
01:10
So from here we can write x is equal to 5 .94 cos omega t and the y coordinate is 5 .94 sine omega t so in the first part of the question we need to find at t is equal to 16.
01:34
7 second the horizontal coordinate of the particle that is x at t is given as 5 .94 cost and this omega value is given as 6 .41 so this is 6 .41 multiplied by 16 .7 and we will get this x of 16 .7 equal to so if we solve this then we will get minus 1 .74 meter.
02:10
So this is the answer for the first part of the question.
02:16
Now, in the second part of the question, we need to find same that is p equal to 16 .7 second y coordinate of the particle.
02:25
So y of t is given as 5 .94 sine.
02:31
Omega value is 6 .41, multiplied by 16 .7 and the unit is meter.
02:40
So again if you saw this y of 16 .7 is equal to 5 .68 meter and this is the answer for the second part of the question.
02:53
Now in the third part of the question we need to find at t equal to 16 .7 second velocity of the particle.
03:00
So in horizontal direction that is vx t will be equal to d x over dd so if you differentiate this position vector so this we will get us 5 .94 omega and cos omega t if you differentiate this so this will become negative of 5 .94 omega sine omega t so now if you find the value of bx at 16 .7 so this will be minus 5 .94 multiplied by 6 .41 multiplied by sine 6 .41 multiplied by 16 .7.
03:47
So the horizontal component of velocity will be equal to so if we solve this then we will get minus 36 .4 meter per second and this is the answer for the third part of the question...