If the velocity $\left( {in\,\,m{s^{ - 1}}} \right)$ of a particle is given by $4\widehat i + 5t\widehat j$, then the magnitude of its acceleration $\left( {in\,\,m{s^{ - 2}}} \right)$ is

Given. $\overrightarrow V = 4\widehat i + 5t\widehat j\,\,m{s^{ - 1}}$ Acceleration, $\overrightarrow a = \frac{{dv}}{{dt}} = d/dt\left( {4\widehat i + 5\widehat j} \right) = 5\widehat j\,\,m{s^{ - 2}}$ $\left| {\overrightarrow a } \right| = {\sqrt {\left( {50} \right)} ^2} = 5.0\,\,m{s^{ - 2}}$

The co-ordinates of a body moving in a planet at any instant of time $t$ are $x = \alpha {t^2}$ and $y = \beta {t^2}$ The speed of the body is

$x = \alpha {t^2} \Rightarrow {v_x} = \frac{{dx}}{{dt}} = 2\alpha t$ $\begin{array}{l} y = \beta {t^2} \Rightarrow {v_y} = \frac{{dy}}{{dt}} = 2\beta t\\ \therefore {\rm{speed}}\,\,V = \sqrt {{v_x}^2 + {v_y}^2} = 2t\sqrt {{\alpha ^2} + {\beta ^2}} \end{array}$

The correct Option is "$2t\sqrt {{\alpha ^2} + {\beta ^2}} $"

A body is moving along the circumference of a circle of radius '$R$' and completes half of the revolution. Then the ratio of its displacement to distance is

Displacement: Distance $\begin{array}{l} = 2R:\pi R\\ = 2:\pi \end{array}$ .

The correct Option is "$2:\pi $"

If a body covers first half of its journey with uniform speed ${V_1}$ and the second half of the journey with uniform speed ${V_2}$ then the average speed is

average speed $ = \frac{{{S_1} + {S_2}}}{{{t_1} + {t_2}}} = \frac{{2{V_1}{V_2}}}{{{V_1} + {V_2}}}$

The correct Option is "$\frac{{2{V_1}{V_2}}}{{{V_1} + {V_2}}}$"

$\frac{{{V_1}{V_2}}}{{{V_1} + {V_2}}}$

A bullet travelling horizontally looses ${\frac{1}{{20}}^{th}}$ of its velocity while piercing a wooden plank. Then the number of such planks required to stop the bullet is

$\frac{1}{n} = \frac{1}{{20}} \Rightarrow n = 20$ No. of planks $ = \frac{{{n^2}}}{{2n - 1}}$

If ${S_n} = 2 + 0.4n$, find initial velocity and acceleration $\left[ {{\rm{in}}\,\,{\rm{units}}} \right]$

${S_n} = \mu + an - \frac{1}{2}a = \left( {u - \frac{1}{2}a} \right) + an$ $\begin{array}{l} {S_n} = 2 + 0.4n\\ 0.4n = an;\,\,\,\,u - \frac{1}{2}a = 2 \end{array}$

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