A particle is moving along the $x-$axis with its coordinate with the time '$t$' given be $\mathrm{x}(\mathrm{t})=10+8 \mathrm{t}-3 \mathrm{t}^{2} .$ Another particle is moving the $y-$axis with its coordinate as a function of time given by $\mathrm{y}(\mathrm{t})=5-8 \mathrm{t}^{3} .$ At $\mathrm{t}=1\; \mathrm{s},$ the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{\mathrm{v}} .$ Then $\mathrm{v}$ (in $\mathrm{m} / \mathrm{s})$ is

The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8t - 5{t^2})$ meter and $x = 6t\, meter$, where $t$ is in second. the angle with the horizontal at which the projectile was projected is

The trajectory of projectile, projected from the ground is given by $y=x-\frac{x^2}{20}$. Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be $...........\,m$

The figure shows the velocity and the acceleration of a point like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum.........$s$ (Given : $a = 4\, m/s^2, v_0 = 40\, m/s, \phi =143^o$)

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

Give explanation of position and displacement vectors for particle moving in a plane by giving suitable equations.