Kinematics MCQ Questions & Answers in Basic Physics | Physics

$${\text{Average}}\,\,{\text{speed}} = \frac{{{\text{ Total distance}}}}{{{\text{ Total time}}}}$$
Let $${t_1},{t_2}$$  be time taken during first-half and second-half respectively.
$$\eqalign{ & {\text{So,}}\,\,{t_1} = \frac{{100}}{{40}}s \cr & {\text{and}}\,\,{t_2} = \frac{{100}}{v}\;s \cr} $$
So, according to average speed formula
$$\eqalign{ & 48 = \frac{{200}}{{\left( {\frac{{100}}{{40}}} \right) + \left( {\frac{{100}}{v}} \right)}} \cr & {\text{or}}\,\,\frac{1}{{40}} + \frac{1}{v} = \frac{2}{{48}} = \frac{1}{{24}} \cr & {\text{or}}\,\,\frac{1}{v} = \frac{2}{{120}} = \frac{1}{{60}} \cr & \Rightarrow v = 60\,km/h \cr} $$

Centripetal acceleration of rotating body is given by
$${a_c} = \frac{{{v^2}}}{r} = \frac{{{r^2}{\omega ^2}}}{r} = r{\omega ^2}\,\,\left( {{\text{as}}\,v = r\omega } \right)$$
Where, $$\omega $$ is angular frequency, but $$\omega = 2\pi v,$$  where $$v$$ is frequency of rotation.
$$\eqalign{ & \therefore {a_c} = r{\left( {2\pi v} \right)^2} = r4{\pi ^2}{v^2} \cr & {\text{Here,}}\,r = 30\;cm = 30 \times {10^{ - 2}}\;m = 0.30\;m \cr & v = 120\,rev/m = \frac{{120}}{{60}}rev/s = 2\,rev/s \cr & \therefore {a_c} = \left( {0.30 \times 4 \times 3.14 \times 3.14 \times 2 \times 2} \right) = 47.4\;m{s^{ - 2}} \cr} $$

Average speed $$ = \frac{s}{{\frac{{\frac{s}{3}}}{{10}} + \frac{{\frac{s}{3}}}{{20}} + \frac{{\frac{s}{3}}}{{60}}}}$$
$$ = \frac{s}{{\frac{s}{{18}}}} = 18\,km/h$$

We have, $$\overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \frac{\pi }{6}$$
$$\eqalign{ & = 4 \times 2 \times \frac{{\sqrt 3 }}{2} = 4\sqrt 3 . \cr & {\text{Now,}}\,{\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + {\left( {\overrightarrow a \cdot \overrightarrow b } \right)^2} = {a^2}{b^2}; \cr & \Rightarrow {\left( {\overrightarrow a \times \overrightarrow b } \right)^2} + 48 = 16 \times 4 \Rightarrow {\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = 16 \cr} $$

As we know, $$R = u\cos \theta \times t$$
Given, $$R = 300\,m, t = 6\,s$$
$$\therefore u\cos \theta = \frac{R}{t} = \frac{{300}}{6} = 50\,m{s^{ - 1}}$$

For $$A$$ to $$B$$
$$S = \frac{1}{2}g{t^2}\,......\left( {\text{i}} \right)$$
For $$A$$ to $$C$$
$$2S = \frac{1}{2}g{{t'}^2}\,......\left( {{\text{ii}}} \right)$$
Dividing (i) by (ii) we get
$$\frac{t}{{t'}} = \frac{1}{{\sqrt 2 }}$$

The distance travelled by the body $$A$$ is $${h_1}$$ given by
$${v_1}t - \frac{{g{t^2}}}{2}$$   and that travelled by the body $$B$$ is $${h_2} = \frac{{g{t^2}}}{2}$$
The distance between the bodies $$ = x = h - \left( {{h_1} + {h_2}} \right).$$
Since $${h_1} + {h_2} = {v_1}t,$$   the relation sought is $$x = h - {v_1}t$$

Initial relative velocity, $${v_{AB}} = v - 0 = v.$$
Acceleration $${a_{AB}} = 0 - a = - a.$$
For max. separation $${v_{AB}} = 0$$
$$0 = {v^2} - 2as \Rightarrow s = \frac{{{v^2}}}{{2a}}$$

$$\eqalign{ & {\text{As}}\,\,\vec C = \vec A + \vec B\,\,{\text{and}}\,\,\left| {\vec C} \right| < \left| {\vec B} \right| \cr & \therefore \left| {\vec C} \right| < \left| {\vec A} \right| \cr} $$

Let the body fall through the height of tower in $$t$$ seconds. From, $${D_n} = u + \frac{a}{2}\left( {2n - 1} \right)$$     we have, total distance travelled in last 2 second of fall is
$$\eqalign{ & D = {D_t} + {D_{\left( {t - 1} \right)}} \cr & = \left[ {0 + \frac{g}{2}\left( {2t - 1} \right)} \right] + \left[ {0 + \frac{g}{2}\left\{ {2\left( {t - 1} \right) - 1} \right\}} \right] \cr & = \frac{g}{2}\left( {2t - 1} \right) + \frac{g}{2}\left( {2t - 3} \right) \cr & = \frac{g}{2}\left( {4t - 4} \right) \cr & = \frac{{10}}{2} \times 4\left( {t - 1} \right) \cr & {\text{or,}}\,\,40 = 20\left( {t - 1} \right)\,\,{\text{or}}\,\,t = 2 + 1 = 3s \cr} $$
Distance travelled in $$t$$ second is
$$s = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times 10 \times {3^2} = 45\,m$$