When two bodies have equal veloci.ties (i.e., \(\overrightarrow { v_{ a } } =\overrightarrow { v_{ b } } =\overrightarrow { v } \)), then their relative velocity is zero i.e., \(\overrightarrow { v_{ ab } } =\overrightarrow { v_{ a } } -\overrightarrow { v_{ b } } =\overrightarrow { v } -\overrightarrow { v } =0.\)

When either the lift is at rest or the lift is moving either vertically upward or downward with a constant speed, we can apply three simple kinetnatic motion equations presuming a = ± g (as the case may be).
In present case u = 49 ms^-1 (upward) a = g = 9.8 ms^-2 (downward)
If the ball returns to boy's hands after a time t, then displacement of ball relative to boy is zero
i.e., s = O. Hence, using equation s = ut + \(\frac{1}{2}\) at², we have
0 = 49 ± - \(\frac {1} {2}\) x 9.8 x t^{2
\(\Rightarrow\)} 4.9 t² - 49t = 0 \(\Rightarrow\) t = 0 or 10 s
As t = 0 is physically not possible, hence time t = 10 s.

Time taken in reaching the office=\(\frac { distance }{ speed } =\frac { 2.5 }{ 5 } =0.5h\) 
Time taken in returning from office\(=\frac { 2.5 }{ 25 } =0.1, \ h=6min\) 

It means that woman reaches the office at 9:30 am and returns home at 5:06 pm.

In the above x-t graph, the slope of x-t graph is increasing, it means that the velocity is increasing.

Instantaneous speed is measured by the speedometer.

As given, x = 2t² + 3t
by differentiating  x w.r.t, we get 
Velocity,    \(v=\frac { dx }{ dt } =\frac { d }{ dt } ({ 2 }t^{ 2 }+3t)\)
v = (4t +3)
As velocity is time dependent, it means that motion is non-uniform

Let x be the displacement . Then,  x = at²
\(\therefore\)      Velocity of the object , v = \(\frac{dx}{dt}\)
Accleration  of the object , a = \(\frac{dv}{dt}\)
It means that a is constant.

Yes, when two bodies move in opposite direction then relative velocity of each is greater than the individual velocities.

Yes,the particle has a uniform acceleration because it does not depend on time t

Given, x = 6 + 18t + 9t²
\({ v }_{ 1 }=\frac { dx }{ dt } =18+18t\)
At t = 2, v₂ = 18 + 18 x 2 = 54 m/s

Any object can be considered as a point object if the distance travelled by it is very large in comparison to its dimensions.
Man along with monkey is cycling smoothly which indicates that the distance travelled by the man is very large, therefore monkey can be taken as a point object.

It is true because if the velocity of the object is negative, then slope of v-t graph is negative.

No, because the size of the beaker is not negligible as compared to the height of the table.

Time taken in downstream,
8 = \(\frac { S }{ { v }_{ r }+v_{ b } } \)
Given v_r + v_b = \(\frac { s }{ 8 } \),v_b-v_r = \(\frac { s }{ 12 } \)
By solving the equations, we get
v_b = \(\frac { s }{ 12 } \left( \frac { 1 }{ 8 } +\frac { 1 }{ 12 } \right) \)
or, v_b = \(\frac { s }{ 2 } \times \frac { 20 }{ 96 } \)
or, v_b = \(\frac { 10s }{ 96 } \)
Now, v_r = \(\frac { 10s }{ 96 } -\frac { S }{ 12 } =\frac { 2S }{ 96 } \)
In still water, only the velocity is to be considered.
∴ time taken in still water for covering length s is,
t = \(\frac { S }{ v_{ b } } =\frac { S\times 96 }{ 10S } \)
= 9.6 seconds.

Relative speed = (34 + 38) kmh^-1 = 72 krnh^-1
= 72 x - ms^-1 = 20 ms^-1
Total distance = (109 + 91) m = 200 m
Time = \(\frac { 200m }{ 20ms^{ -1 } } \) =10 s.

No. because the x-t graph does not represent the trajectory of the path followed by a particle. From the graph, it is noted that at t = 0, x =0. Context The abovb graph can represent the motion of a body falling freely from a tower under gravity.

In the SHM, acceleration a = -aix, where m(i.e. angular frequency) is constant.
(i) At time t = 0.3s, x is negative, the slope of x-t plot is negative, hence position and velocity are negative. Since \(a={ -\omega }^{ 2 }x\), hence acceleration is positive.
(ii) At time t = 1. 2s, x is positive, the slope of x-t plot is also positive, hence position and velocity are positive. Since \(a={ -\omega }^{ 2 }x\), hence acceleration is negative.
(iii) At t = -1.2s, x is negative, the slope of x-t plot is also negative. But since both x and t are negative here, hence velocity is positive. Finally, acceleration a is also positive.

(a) No, graph (a) is not representing one-dimensional motion of a particle, because graph shows two different positions of the particle at same instant of time. (At time t₁ particle is at positions P and Q and at time t₂, particle is at positions Rand S), which is not possible.
  
(b) No, graph (b) cannot represent one-dimensional motion of a particle, because graph shows one positive velocity (v₁ ) and another negative velocity (-v₂) of the particle at the same instant of time (t₁) which is not possible.
  
(c) No, graph (c) cannot represent one-dimensional motion of a particle, because graph shows negative speed of the particle but speed cannot be negative.
  
(d) No, graph (d) cannot represent one-dimensional motion of a particle, because graph shows that total path length increases from time t = 0 to t = t₁, but decreases from t = t₁ to t = t₂ But total path length of a moving particle can never decrease with time.
   

Instantaneous speed (v_ins) of the particle at an instant is the first derivative of the distance with respect to time at that instant of time i.e. \({ v }_{ ins }=\frac { dx }{ dt } .\)
Since, in instantaneous speed, we take only a small interval of ime(dt) during which direction of motion of a body is not supposed to change, hence there is nodifference between total path length and magnitude of displacement for small interval of time dt.
Hence, instantaneous speed is always equal to magnitude of instantaneous velocity.

To fully cross the bridge, distance to covered by end far away from train is equal to (500+1000) m
\(\text { Average speed } -\frac{1500}{10} =150 \mathrm{~m} / \mathrm{s}\)

(i) The given equation x = at² - bt³
If t = 0, x_o=0
if t = 2s,x₂ = 4a - 8b
\(\triangle x={ x }_{ 2 }-{ x }_{ 0 }=4a-8b-0=4a-8b\)
Average speed in the given interval of time.
\({ v }_{ av }=\frac { \triangle x }{ \triangle t } =\frac { 4a-8b }{ 2 } =2a-4b\)
(ii) Instantaneous speed
\(v=\frac { dx }{ dt } =\frac { d }{ dt } ({ at }^{ 2 }-{ bt }^{ 3 })=2at-3b{ t }^{ 2 }\)
At t = 2s, v = 4a - 12 b m/s

(i) False, because velocity is the speed of body in a given direction. When speed is zero, the magnitude of velocity of body is zero, hence velocity is zero.
(ii) True, when a particle is moving along a straight line with a constant speed, its velocity remains constant with time. Therefore, acceleration (i,e. change in velocity/time) is zero.
(iii) False, if the initial velocity of a body is negative, then even in the case of positive acceleration, the body speeds down. A body speeds up when the acceleration acts in the direction of motion.

(i) Since, the ball is moving under the effect of gravity, the direction of acceleration due to gravity is always vertically downwards.
(ii) At the highest point, the velocity of the ball becomes zero and acceleration is equal to the acceleration due to gravity=9.8ms^-2 in vertically downward direction.
(iii) When the highest point is chosen as the location for x=0 and t=0 and vertically downward direction to be the positive direction of x-axis and upward direction as negative direction of x-axis.
During upward motion, sign of position is negative, sign of velocity is negative and sign of acceleration is positive.During downward motion, sign of position is positive, sign of velocity is positive and sign of acceleration is also positive.
(iv) Let t be the time taken by the ball to reach the highest point where height from ground be s.
Taking vertical upward motion of the ball, we have u  = -29.4ms^-, a = 9.8ms^-, v = 0, s = S,t = ?
As, v² - u² = 2 as
0 - (-29.4)²=2 x 9.8 x S
or \(S=\frac { -{ (29.4) }^{ 2 } }{ 2\times 9.8 } =-44.1m\)
Here, negative sign shows that the distance is covered in upward direction.
As, v = u + at
\(\therefore 0=-29.4+9.8\times t\ or\ t=\frac { 29.4 }{ 9.8 } =3s\)
It means time of ascent = 3s
When an object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
Therefore, total time after which the ball returns to the player's hand = 3 + 3 = 6s.

\( \frac{dv}{dt}=-kv^2\)
Integrating both sides, we get
\( \int\frac{dv}{v^3}=-k\int dt\)
or \( -\frac{1}{2v^2}=-kt+c\)
At t = 0,v = v_0
∴ c= \(-\frac{1}{2v_0^2}\)
∴ \(-\frac{1}{2v^2}=-kt-\frac{1}{v_0^2}\)
or \(2v^2=\frac{2v_0^2}{(2v_0^2 kt+1)}\)
or v= \(\sqrt{\frac{v_0^2}{(2v_0^2 kt+1)}}\)

(iii) What will be the position-time curve for dropped packet?
In position-time curve, when a packet is dropped. Let us consider origin at the point where packet was dropped.

Let us consider left to right to be the positive direction of x-axis.
(i) Here, velocity of belt, VB = + 4 kmh^-1,speed of child w.r.t. belt Vc = + 9 kmh^-1 = 5/2 ms^-1
Speed of the child w.r.t. stationary observer,
V'_C = v_C + v_B= 9 + 4 = 13kmh^-1
(ii) Here, vB = + 4 kmh -1, "c = - 9 kmh "
Speed of the child w.r.t. stationary observer,
v'_C = v_C + v_B = - 9 + 4 = - 5 krnh^-1
Here, negative sign shows that the child will appear to run in a direction opposite to the direction of motion of the belt.
(iii) Distance between the parents, s = 50 m Since parents and child are located on the same belt, the speed of the child as observed by stationary observer in either direction (either from mother to father or from father to mother) will be 9 km -I Time taken by child in case (i) and (ii) is
\(t=\frac { 50 }{ (5/2) } =20\)
If motion is observed by one of the parents, answer to case (i) or case (ii) will get altered. It is so because speed of child w.r,t either of mother or father is 9 kmh^-1 But answer (iii) remains unaltered due to the fact that parents and child are on the same belt and as such all are equally affected by the motion of the belt.

The slope of the given graph over the time interval t₁, to t₂ is not constant and is not uniform. It means acceleration is not constant or uniform, therefore relations (i), (ii) and (v) are not correct which is uniform accelerated motion, but relations (iii), (iv) and (vi) are correct, because these relations are true for both uniform or non-uniform accelerated motion.