The position vector of a particle `vec(R )` as a funtion of time is given by:
`vec(R )= 4sin(2pit)hat(i)+4cos(2pit)hat(j)`
Where `R` is in meters, `t` is in seconds and `hat(i)` and `hat(j)` denote until vectors along x-and y- directions, respectively Which one of the following statements is wrong for the motion of particle ?

Options

A. Acceleraion is along -R
B. Magnitude of accleration vector is `(v^(2))/(R ).` where v in the velocity of particle is 8 m/s
C. path of the particle is a circle of radius 4 m
D. magnitude of the velocity of particle is `8m//s`

Answer

c

Explanation

The position vector of a particle R as a
function of time is given by
`R=4sin(2pit)hati+4cos(2pit)hatj`
x- axis component, x = 4 `sin2pit`
y-axis component,y = 4 `cos2pit`
Squaring and adding both equations we get
`x^(2)+y^(2)=4^(20(sin^(2)(2pit)+cos^(2)(2pit)`
i.e `x^(2)+y^(2)=4^(2)`i.e equations of circle and radius is 4 m
Acceleration vectors, `a=(v^(2)/(R)(-hatR0.` while v is velocity of a particle
Magnitude of acceleration vector `a=(v^(2))/(R)`
As, we have `v_(x)=+4(cos2pit)2pi`and
`v_(y)=-4(sin2pit)2pi`
Net resultant velocity
`v=sqrt(v_(x)^(2)+v_(y)^(2))`
`=sqrt((8pi)^(2)(cos^(2)2pit+sin^(2)2pit))`
`v=8pi`
`| :.cos^(2)2pit+sin^(2)2pit=1|`
So, option (c0 is incorrect