Kinematics MCQ Questions & Answers in Basic Physics | Physics

Here, initial velocity of ball $$u = 40\,m/s$$
Acceleration of ball $$a = - g\,m/{s^2} = - 10\,m/{s^2}$$
Time $$= 2s$$
From first equation of motion,
$$\eqalign{ & v = u + at \cr & v = 40 - 10 \times 2 \cr & \Rightarrow v = 20\,m/s \cr} $$

Velocity of $$A$$ is same as that of $$B$$ in magnitude but opposite in direction.

$$\eqalign{ & h = ut - \frac{1}{2}g{t^2} \cr & \Rightarrow g{t^2} - 2ut + 2h = 0 \cr} $$
solving for $$t$$ we get
$$\eqalign{ & {t_1} + {t_2} = \frac{{2u}}{g} \cr & {t_1} \times {t_2} = \frac{{2h}}{g} \cr & {\text{so,}}\,\Delta t = \left| {{t_1} - {t_2}} \right| = {\left( {{t_1} + {t_2}} \right)^2} - 4{t_1}{t_2} \cr} $$
putting value we get
$$u = \frac{1}{2}\sqrt {8gh + {g^2}\Delta {t^2}} $$

Only option (B) is false since acceleration vector is always radial (i.e., towards the center) for uniform circular motion.

When the wheel rolls on the ground without slipping and completes half rotation, point $$P$$ takes new position as $$P’$$ as shown in figure.

Horizontal displacement, $$x = \pi R$$
Vertical displacement, $$y = 2R$$
Thus, displacement of the point $$P$$ when wheel completes half rotation,
$$\eqalign{ & s = \sqrt {{x^2} + {y^2}} \cr & = \sqrt {{{\left( {\pi R} \right)}^2} + {{\left( {2R} \right)}^2}} \cr & = \sqrt {{\pi ^2}{R^2} + 4{R^2}} \cr & {\text{but}}\,R = 1\,m\,\,\left( {{\text{given}}} \right) \cr & \therefore s = \sqrt {{\pi ^2}{{\left( 1 \right)}^2} + 4{{\left( 1 \right)}^2}} \cr & = \sqrt {{\pi ^2} + 4} \,m \cr} $$

Velocity of $$A$$ w.r.t. $$B$$ is given by $${v_{AB}} = {v_A} - {v_B}.$$
Relative velocity of the parrot w.r.t. the train $$ = \left[ {10 - \left( { - 5} \right)} \right]m{s^{ - 1}}$$
$$ = 15\,m{s^{ - 1}}.$$
Time taken by the parrot to cross the train $$ = \frac{{150}}{{15}} = 10\,s$$

Let $$'S'$$ be the distance between two ends $$'a'$$ be the constant acceleration.
As we know $${v^2} - {u^2} = 2aS\,\,{\text{or,}}\,\,aS = \frac{{{v^2} - {u^2}}}{2}.$$
Let $$v$$ be velocity at mid point.
Therefore, $$v_c^2 - {u^2} = 2a\frac{S}{2} \Rightarrow v_c^2 = {u^2} + aS$$
$$v_c^2 = {u^2} + \frac{{{v^2} - {u^2}}}{2} \Rightarrow {v_c} = \sqrt {\frac{{{u^2} + {v^2}}}{2}} $$

Let the man starts crossing the road at an angle $$\theta $$ as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance $$4 + AC$$  or $$4 + 2\cot \theta .$$
$$\therefore \frac{{4 + 2\cot \theta }}{8} = \frac{{\frac{2}{{\sin \theta }}}}{v}\,\,{\text{or}}\,\,v = \frac{8}{{2\sin \theta + \cos \theta }}\,......\left( {\text{i}} \right)$$
For minimum $$v,\frac{{dv}}{{d\theta }} = 0$$

$$\eqalign{ & {\text{or}}\,\,\frac{{ - 8\left( {2\cos \theta - \sin \theta } \right)}}{{\left( {2\cos \theta + \sin \theta } \right)}} = 0 \cr & {\text{or}}\,2\cos \theta - \sin \theta = 0\,\,{\text{or}}\,\,\tan \theta = 2 \cr} $$
From equation (i),
$${v_{\min }} = \frac{8}{{2\left( {\frac{2}{{\sqrt 5 }}} \right) + \frac{1}{{\sqrt 5 }}}} = \frac{8}{{\sqrt 5 }} = 3.57\,m/s$$

Average speed can be calculated as the total distance travelled divided by the total time taken.

Let $${t_1},{t_2},{t_3}$$  be times taken in covering distances $$PR, RS$$  and $$SQ$$  respectively.
$$\eqalign{ & \therefore {t_1} = \frac{{\left( {\frac{s}{3}} \right)}}{{10}}, \cr & {t_2} = \frac{{\left( {\frac{s}{3}} \right)}}{{20}} \cr & {\text{and}}\,\,{t_3} = \frac{{\left( {\frac{s}{3}} \right)}}{{60}} \cr & \therefore {\text{Average speed}} = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{s}{{{t_1} + {t_2} + {t_3}}} \cr & = \frac{s}{{\frac{{\left( {\frac{s}{3}} \right)}}{{10}} + \frac{{\left( {\frac{s}{3}} \right)}}{{20}} + \frac{{\left( {\frac{s}{3}} \right)}}{{60}}}} \cr & = \frac{s}{{\left( {\frac{s}{{18}}} \right)}} \cr & = 18\,km/h \cr} $$

Concept
Average velocity is defined as the ratio of displacement to time taken while the average speed of a particle in a given interval of time is defined as the ratio of distance travelled to the time taken.
On a circular path in completing one turn, the distance travelled is $$2\pi r$$  while displacement is zero.
Hence, $${\text{average velocity}} = \frac{{{\text{displacement}}}}{{{\text{Time - interval}}}}$$
$$ = \frac{0}{t} = 0$$
$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Distance}}}}{{{\text{Time - interval}}}} \cr & = \frac{{2\pi r}}{t} = \frac{{2 \times 3.14 \times 100}}{{62.8}} \cr & = 10\,m{s^{ - 1}} \cr} $$
NOTE
If a particle moves in a straight line without change in direction, the magnitude of displacement is equal to the distance travelled otherwise it is always less than it. Thus, displacement $$ \leqslant $$ distance.