Question

In a curvilinear motion particle P moves along the fixed path 9y=x^2,where x and y are expressed in cm.At any instant t, the x coordinate of P is given by x=t^2-14t.Determine y component of velocity and acceleration of P when t=15 sec?

Answer 1

$$g_y = x^2; \quad x(t) = t^2 - 14t; \quad t = 15 \sec$$

The position of a particle P in two-dimensional Cartesian $(x, y)$ coordinates, with respect to time

$$\begin{aligned} x_p(t) &= t^2 - 14t & \quad & \quad & (cm) \\ y_p(t) &= \frac{x^2}{y} &= \frac{(t^2 - 14t)^2}{y} & \quad & (cm) \end{aligned}$$

The velocity of the particle P is given by:

$$\begin{aligned} v_x(t) &= \frac{dv_p(t)}{dt} = 2t - 14 & \quad & \quad & (cm/\sec) \\ v_y(t) &= \frac{dv_p(t)}{dt} = \frac{2}{y}(t^2 - 14t)(2t - 14) & \quad & \quad & (cm/\sec) \end{aligned}$$

The acceleration of the particle P is given by:

$$\begin{aligned} a_x(t) &= \frac{dv_x(t)}{dt} = 2 & \quad & \quad & (cm/\sec^2) \\ a_y(t) &= \frac{dv_y(t)}{dt} = \frac{2}{y}[(2t - 14)^2 + 2t^2 - 28t] & \quad & \quad & (cm/\sec^2) \end{aligned}$$

The magnitude of the velocity of particle P:

$$\begin{aligned} v_p(t) &= \sqrt{(v_x(t))^2 + (v_y(t))^2} = \\ &= \sqrt{(2t - 14)^2 + \left(\frac{2}{y}(t^2 - 14t) \cdot (2t - 14)\right)^2} \end{aligned}$$

at t = 15

$$V_p(15) = \sqrt{(2.15 - 14)^2 + \left(\frac{2}{9}(15^2 - 14.15)(2.15 - 14)\right)^2}$$ $$= 55.7 \text{ cm/sec}$$

The magnitude of the acceleration of the particle $P$:

$$a_p(t) = \sqrt{(a_x(t))^2 + (a_y(t))^2}$$ $$= \sqrt{2^2 + \left(\frac{2}{9}[(2t - 14)^2 + 2t^2 - 28t]\right)^2}$$

at t = 15

$$a_p(15) = \sqrt{4 + \left(\frac{2}{9}[(2.15 - 14)^2 + 2.15^2 - 28.15]\right)^2}$$ $$= 63.6 \text{ cm/sec}^2$$