Kinematics MCQ Questions & Answers in Basic Physics | Physics

First of all find acceleration from the given values and then using equation of motion calculate distance travelled.
Given, Initial velocity $$u = 0,$$   time $$t = 20\,s$$
Final velocity $$v = 144\,km/h$$
$$= 40\,m/s$$
From 1st equation of motion,
$$\eqalign{ & v = u + at \cr & \Rightarrow a = \frac{{v - u}}{t} = \frac{{40 - 0}}{{20}} = 2\,m/{s^2} \cr} $$
Now, from 2nd equation of motion, distance covered, $$s = ut + \frac{1}{2}a{t^2} = 0 + \frac{1}{2} \times 2 \times {\left( {20} \right)^2}$$
$$ = 400\,m$$
NOTE
If the car decelerates, the acceleration will be negative and is termed as retardation.

$$\eqalign{ & {\text{We know that,}} \cr & v = \frac{{dx}}{{dt}} \Rightarrow dx = v\,dt \cr & {\text{Integrating}},\int\limits_0^x {dx} = \int\limits_0^t {v\,dt} \cr & {\text{or}},\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)dt = \left[ {{v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3}} \right]_0^t} \cr & {\text{or}},\,x = {v_0}t + \frac{{g{t^2}}}{2} + \frac{{f{t^3}}}{3} \cr & {\text{At }}t = 1,\,\,\,x = {v_0} + \frac{g}{2} + \frac{f}{3}. \cr} $$

Velocity of boat $$ = \frac{{8 + 8}}{2} = 8\,km\,{h^{ - 1}}$$
Velocity of water $$ = 4\,km\,{h^{ - 1}}$$
$$t = \frac{8}{{8 - 4}} + \frac{8}{{8 + 4}} = \frac{8}{3}h = 160\,{\text{minute}}$$

$$\eqalign{ & {a_c} = \frac{{{v^2}}}{r} = \frac{{{{\left( {250} \right)}^2}}}{{{{10}^3}}} = 62.5\,m/{s^2} \cr & \Rightarrow \frac{{{a_c}}}{g} = \frac{{62.5}}{{9.8}} = 6.38 \cr} $$

$$\eqalign{ & {y_1} = 10t - 5{t^2};\,\,{y_2} = 40t - 5{t^2} \cr & {\text{for}}\,{y_1} = - 240\,m,\,\,t = 8\,s \cr & \therefore {y_2} - {y_1} = 30\,t\,{\text{ for }}t \leqslant 8\,s \cr & {\text{for }}t > 8\,s, \cr & {y_2} - {y_1} = 240 - 40\,t - \frac{1}{2}g{t^2} \cr} $$

Using $${v^2} - {u^2} = 2\,as$$
$${\left( {20} \right)^2} - {\left( {10} \right)^2} = 2 \times a \times 135 \Rightarrow \frac{{300}}{{270}} = a = \frac{{10}}{9}$$
Now, using $$v - u = at$$
$$\eqalign{ & 20 - 10 = \frac{{10}}{9} \times t \cr & \Rightarrow t = 9\,s \cr} $$

Concept
If two vectors are perpendicular to each other than their dot product is always equal to zero.
$$\eqalign{ & {\text{Let,}}\,a = 2\hat i + 3\hat j + 8\hat k \cr & b = 4\hat j - 4\hat i + \alpha \hat k \cr & = - 4\hat i + 4\hat j + \alpha \hat k \cr} $$
According to the above hypothesis
$$\eqalign{ & a \bot b \cr & \Rightarrow a \cdot b = 0 \cr & \Rightarrow \left( {2\hat i + 3\hat j + 8\hat k} \right) \cdot \left( { - 4\hat i + 4\hat j + \alpha \hat k} \right) = 0 \cr & \Rightarrow - 8 + 12 + 8\alpha = 0 \cr & \Rightarrow 8\alpha = - 4 \cr & \therefore \alpha = - \frac{4}{8} \cr & = - \frac{1}{2} \cr} $$
NOTE
$$a \cdot b = ab\cos \theta .$$    Here, $$a$$ and $$b$$ are always positive as they are the magnitudes of $$a$$ and $$b.$$

$$t = \frac{{2u\sin \left( {\alpha - \beta } \right)}}{{g\cos \beta }};\,u = \frac{{gt\cos \beta }}{{2\sin \left( {\alpha - \beta } \right)}}$$

We have $$\overrightarrow a \times \overrightarrow b = 39\overrightarrow k = \overrightarrow c $$
Also $$\left| {\overrightarrow a } \right| = \sqrt {34} ,\left| {\overrightarrow b } \right| = \sqrt {45} ,\left| {\overrightarrow c } \right| = 39;$$
$$\therefore \left| {\overrightarrow a } \right|:\left| {\overrightarrow b } \right|:\left| {\overrightarrow c } \right| = \sqrt {34} :\sqrt {45} :39.$$

$$\eqalign{ & \hat u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \cr & \Rightarrow u\,\cos \theta = 1,\,\,\,\,u\,\sin \theta = 2 \cr & y = x\,\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}} \cr & \therefore y = 2x - \frac{1}{2}g{x^2} = 2x - 5{x^2} \cr} $$