A person walks $25.0^{\circ}$ north of east for $3.18 \,km$. How far would she have to walk due north and then due east to arrive at the same location?

A car starts from rest and accelerates at $5 \,\mathrm{~m} / \mathrm{s}^{2}$. At $t=4 \mathrm{~s}$, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $\mathrm{t}=6\, \mathrm{~s}$ ?

(Take g $\left.=10\, \mathrm{~m} / \mathrm{s}^{2}\right)$

The initial velocity of a projectile is $\vec u = (4\hat i + 3\hat j)\,m/s$ it is moving with uniform acceleration $\vec a = (0.4\hat i + 0.3\hat j)\, m/s^2$ The magnitude of its velocity after $10\,s$ is………$m/s$

If vectors $\overrightarrow {A} = cos\omega t\hat i + sin\omega t\hat j$ and $\overrightarrow {B} = cos\frac{{\omega t}}{2}\hat i + sin\frac{{\omega t}}{2}\hat j$ are functions of time, then the value of $t$ at which they are orthogonal to each other is 

The $x$ and $y$ coordinates of the particle at any time are $x = 5t – 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The acceleration of the particle at $t = 2\, s$ is……$m/sec^2$