Answer to Question #179852 in Math for Nii Laryea

Question

A particle is moving in the ~i - ~j plane with an acceleration of (1 - t)~ i+ t ~j where t is the time. The particle is projected from the point with an initial velocity of 3/2 ~i ms/1. Find the displacement of the particle at time t and the value of t when it is moving in the direction of motion.

What is the angle between its velocity and displacement at this time?

Answer 1

$$\vec a(t)=(1-t)\vec i+t\vec j$$

$$\vec v(t)=(t-\dfrac{1}{2}t^2)\vec i+\dfrac{1}{2}t^2 \vec j+\vec v_0$$

$$=(t-\dfrac{1}{2}t^2+\dfrac{3}{2})\vec i+\dfrac{1}{2}t^2 \vec j$$

$$\vec r(t)=(\dfrac{1}{2}t^2-\dfrac{1}{6}t^3+\dfrac{3}{2}t)\vec i+\dfrac{1}{6}t^3 \vec j+\vec r_0$$

$$\Delta \vec r=(\dfrac{1}{2}t^2-\dfrac{1}{6}t^3+\dfrac{3}{2}t)\vec i+\dfrac{1}{6}t^3 \vec j$$

$$\tan\theta=\dfrac{\dfrac{1}{6}t^3}{\dfrac{1}{2}t^2-\dfrac{1}{6}t^3+\dfrac{3}{2}t}$$

$$\tan\varphi=\dfrac{\dfrac{1}{2}t^2}{t-\dfrac{1}{2}t^2+\dfrac{3}{2}}$$

Let $\tan\theta=\tan\varphi.$ Then

$$\dfrac{\dfrac{1}{6}t^3}{\dfrac{1}{2}t^2-\dfrac{1}{6}t^3+\dfrac{3}{2}t}=\dfrac{\dfrac{1}{2}t^2}{t-\dfrac{1}{2}t^2+\dfrac{3}{2}}$$

$$2t^4-t^5+3t^3=3t^4-t^5+9t^3$$

$$t^4+6t^3=0$$

Since $t>0,$ then there is no solution.

When particle is moving in the direction of motion the angle between its velocity and displacement is zero.