The particle’s velocity function will have the form v(t) = ntdi + otej + ptgk. enter an expression for d in terms of x, y, z, a, b, and c. 25% part (c) here is a set of parameter values for the motion of the particle: m = 1.8 kg, x = 1.5 m/s3, y = 2.8 m/s4, z = 0.15 m/s3, a = 3, b = 4, c = 3. calculate the x-component of the particle’s angular momentum, in units of kg˙m2/s, about the origin at time t = 1 s. 25% part (d) use the same set of parameter values (m = 1.8 kg, x = 1.5 m/s3, y = 2.8 m/s4, z = 0.15 m/s3, a = 3, b = 4, c = 3) to calculate the y-component of the particle’s angular momentum, in units of kg˙m2/s, about the origin at time t = 1 s.

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction