Problem 1: The position vector of a point is given by, \vec{r}=(\frac{4}{3}t^3-2t)\hat{i}+t^2\hat{j} Find the velocity and acceleration of the point at t=3sec. [The distance is measured in meters]
Problem 2: The motion of a particle is described by the equation, x=4\sin{2t} , y=4\cos{2t} and z=6t . Find the equation of velocity and acceleration of the particle.
Problem 3: A particle moves along a curve x=2\sin{3t} , y=2\cos{3t} and z=8t . At any instant of time (t>0), find the velocity and acceleration of the particle.
Problem 4: The velocity of a moving particle at any instant of time is given by, \displaystyle{\vec{v}=2\hat{i}+5t\hat{j}+\frac{1}{t}\hat{k}} Find the position vector of that particle at that instant of time.
Problem 5: A particle is moving along a curve in a plane. Derive an expression for the displacement, radial and transverse components of velocity and acceleration. Prove that for the motion of a particle in a plane, \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta} and \vec{a}=(\ddot{r}-r\ {\dot{\theta}}^2)\hat{r}+(\ddot{r}\theta+2r\ddot{\theta})\hat{\theta}
Problem 8: The path of a projectile is defined by the equation \displaystyle{r=3t-\frac{1}{30}t^2} and {\theta}^2=1600-t^2 . Find its velocity and acceleration after 30 sec.
Problem 9: For planar motion, x=r\ \cos\theta and y=r\ \sin\theta . Prove that \displaystyle{\dot{r}=\frac{x\dot{x}+y\dot{y}}{r}} and, \displaystyle{r\dot{\theta}=\frac{x\dot{y}-y\dot{x}}{r}}
Problem 10: A point moving in a plane has co-ordinates x=3 and y=4. And has components of speed \dot{x}=5 m/sec and \dot{y}=8 m/sec at some instant of time. Find the components of speed in polar co-ordinates r , \theta along the direction \hat{r} and \hat{\theta} respectively.
The polar co-ordinates of a particle moving in a plane are given by, r=a\sin{\omega}_1{t} and \theta={\omega}_2{t} . Obtain an expression for the polar components of the velocity and acceleration of the particle.
Show that the solid angle subtended by a ring element cut from a sphere of radius R is given by, d\omega=4\pi\ \sin\frac{\theta}{2}\ \cos\frac{\theta}{2}\ d\theta , where \theta is the angle between the normal through the centre of the ring and the line joining centre of the sphere with a point on the internal circumference and d\theta is the angular width of the element.