The position vector r of a particle moving in the xy plane is r=2ti+2sin[(pi/4)t]j , with is in meters and t in seconds. (a) Calculate the x and y components of the particle's position at , and 4.0 s and sketch the particle's path in the plane for the interval . (b) Calculate the components of the particle's velocity at , and 3.0 s. Show that the velocity is tangent to the path of the particle and in the direction the particle is moving at each time by drawing the velocity vectors on the plot of the particle's path in part (a). (c) Calculate the components of the particle's acceleration at , and 3.0 s.

a.

b. velocity= dr/dt
showing it is tangent is to show the slope is the negative inverse, that is, the i component of one is the -j

c. acceleration=dv/dt

(a) To calculate the x and y components of the particle's position at different times, we can substitute the given value of t into the position vector r.

At t = 2.0 s:
Substituting t = 2.0 s into the position vector:
r = 2(2.0)i + 2sin[(π/4)(2.0)]j
= 4.0i + 2sin[(π/2)]j
= 4.0i + 2j

So, the x-component is 4.0 m and the y-component is 2.0 m.

At t = 4.0 s:
Substituting t = 4.0 s into the position vector:
r = 2(4.0)i + 2sin[(π/4)(4.0)]j
= 8.0i + 2sin[(π)]j
= 8.0i

So, the x-component is 8.0 m and the y-component is 0.0 m.

To sketch the particle's path in the xy-plane, plot the points (4.0, 2.0), and (8.0, 0.0) on the graph. Connect these points with a smooth curve. The resulting path will be a portion of a sinusoidal curve.

(b) To calculate the components of the particle's velocity at different times, we differentiate the position vector r with respect to time (t).

Velocity vector v = d/dt (2ti + 2sin[(π/4)t]j)
= 2i + 2cos[(π/4)t](π/4)j

At t = 2.0 s:
Substituting t = 2.0 s into the velocity vector:
v = 2i + 2cos[(π/4)(2.0)](π/4)j
= 2i - 2(π/4)j
= 2i - (π/2)j

So, the x-component of the velocity is 2.0 m/s and the y-component is -(π/2) m/s.

At t = 3.0 s:
Substituting t = 3.0 s into the velocity vector:
v = 2i + 2cos[(π/4)(3.0)](π/4)j
= 2i + 2cos[(3π/4)](π/4)j

To find the value of cos[(3π/4)], we use the symmetry property of cosine function:
cos[(3π/4)] = -cos[(π/4)]

v = 2i + 2(-cos[(π/4)](π/4)j
= 2i - 2cos[(π/4)](π/4)j
= 2i - 2(π/4)j
= 2i - (π/2)j

So, the x-component of the velocity is 2.0 m/s and the y-component is -(π/2) m/s.

To show that the velocity vector is tangent to the path of the particle and in the direction the particle is moving at each time, we draw the velocity vectors (2.0, -(π/2)) at t = 2.0 s and t = 3.0 s on the graph of the particle's path from part (a). The velocity vectors should be drawn from the corresponding points on the path in the direction indicated by the vector components.

(c) To calculate the components of the particle's acceleration, we differentiate the velocity vector v with respect to time (t).

Acceleration vector a = d/dt (2i + 2cos[(π/4)t](π/4)j)
= 0i - 2sin[(π/4)t](π/4)^2j

At t = 2.0 s:
Substituting t = 2.0 s into the acceleration vector:
a = 0i - 2sin[(π/4)(2.0)](π/4)^2j
= 0i - 2sin[(π/2)](π/16)j
= 0i - 2(1)(π/16)j
= 0i - (π/8)j

So, the x-component of the acceleration is 0.0 m/s^2 and the y-component is -(π/8) m/s^2.

At t = 3.0 s:
Substituting t = 3.0 s into the acceleration vector:
a = 0i - 2sin[(π/4)(3.0)](π/4)^2j
= 0i - 2sin[(3π/4)](π/16)j

To find the value of sin[(3π/4)], we use the symmetry property of sine function:
sin[(3π/4)] = -sin[(π/4)]

a = 0i - 2(-sin[(π/4)])(π/16)j
= 0i + 2sin[(π/4)](π/16)j
= 0i + 2(1)(π/16)j
= 0i + (π/8)j

So, the x-component of the acceleration is 0.0 m/s^2 and the y-component is (π/8) m/s^2.