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 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.
Problem 11: The polar co-ordinates of a point \( (r,\ \theta,\ \phi) \) = \( (8,\ 30^{\circ},\ 45^{\circ}) \). Find the Cartesian co-ordinates of the same point.
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.