Question

A particle has its position vector defined by

r =(4t−t ) ²+(3−2t² ) 
, with r and t are measured in SI units. At the time when x-component of the particle's velocity is zero, what is the magnitude of y-component of its velocity?

Answer 1

When the x-component of the particle's velocity is zero, the magnitude of the y-component of the particle's velocity is 8 m/s.

Step-by-step explanation:

The student has asked about the velocity components of a particle with a given position vector. To find the magnitude of the y-component of the particle's velocity when the x-component of the velocity is zero, we need to differentiate the position vector, find when the x-component of velocity is zero, and then determine the y-component at that instant.

The position vector is given as r(t) = (4t−t²)î+(3−2t²)ê. The velocity vector v(t) is the derivative of the position vector with respect to time t. Therefore, v(t) = dr(t)/dt = (4−2t)î + (-4t)ê.

To find when the x-component of velocity is zero, we set the x-component of v(t) to zero and solve for t, which gives us 4 - 2t = 0, hence t = 2 seconds. Substituting t into the y-component of v(t), we get vy = -4(2) = -8 m/s. Therefore, the magnitude of the y-component of velocity when the x-component is zero is 8 m/s.