Explanation
The average speed of an object is defined as the distance traveled divided by the time
$$v=\dfrac{D}{t}$$
Where, $$v$$ = speed,
$$ D$$ = distance and
$$t$$= time
Let the initial velocity$$u=0$$
The time taken and the distance traveled during accelerated motion and retarding motion will be same as the acceleration is same
For accelerated motion
$$v=u+at=5t$$
And for retardation
$$0=v-5t$$
$$v=5t$$
As the velocity v will become u for retardation
Total distance traveled
$$=$$ area of $$v-t$$
$$=$$ area$$I+$$ area $$II+$$ area $$III$$
$$2\left( \cfrac { 1 }{ 2 } 5{ t }^{ 2 } \right) +5tx=$$distance traveled during motion
$$x=$$time during which particle remains in uniform motion
$$5{ t }^{ 2 }+5tx=$$speed$$x$$ time
$$5{ t }^{ 2 }+5tx=20\times 25$$
$${ t }^{ 2 }+tx=100\rightarrow 1$$
Also the total time is given $$25$$seconds
Thus $$2t+x=25$$
Thus $$x=25-2t\rightarrow 2$$
Substituting $$2$$ in $$1$$
$${ t }^{ 2 }+t(25-2t)=100$$
$${ t }^{ 2 }-25t+100=0$$
$$t=5$$ or $$t=20$$
Thus the time will be $$t=5$$sec as $$t=20$$s is not possible as it will exceed the given time
We know,
The slope of the $$x-t$$ graph represents the speed. The average speed of a particle shown in the $$x-t$$ graph is obtained from its slope in a particular interval of time. So, the higher the slope of the graph, the higher is the average speed.
Now see the graph,
It is clear from the graph that the magnitude of the slope is more in interval-$$1$$ than in the interval-$$2$$.
Therefore, the speed of the particle in $$2$$ will be less than the $$1$$.
Thus $$v_{1} > v_{2}$$
The slope of the Displacement- time graph provides the value of the velocity of a moving object. From the graph, it is visible that, till time $$t_{0}$$ object is having a constant slope and hence its velocity will be constant. Whereas after $$t_{0}$$ slope is zero and hence the object is not moving at all. Thus option D is most suitable and hence is the correct option.
The correct option is D.
$$\textbf{Explanation:}$$
$$\textbf{Step 1: Positions of their respective houses from school}$$
Since the slope of OP is lesser OQ, it infers that point P is closer to O than Q.
⸫ A lives closer to a school than B. Hence, (a) is true.
$$\textbf{Step 2: Their starting times from school}$$
From the graph, for line OP, x=0; t=0
For line OQ, when x=0; t has a positive finite value which means that A started from school earlier than B.
Hence, (b) is also true.
$$\textbf{Step 3: Their respective reaching times}$$
Point P and Q lie at the same vertical line on the abscissa (t) which denotes that both A and B reach their respective homes at the same time ‘t’.
Hence, (d) is also true
$$\textbf{Step 4: Their respective walking speeds}$$
The pace of the children is indicated by the slope of the x-t graph as their motion is uniform motion. From the graph, it can be seen that the slope of OQ is greater than OP which means that pace of B is greater than A.
Hence, (c) is false.
Final Answer : $$\textbf{Hence, the correct option is (c) A walks faster B}$$
For the V-t graph shown displacement during 15 seconds is
Correct Answer: Option (D)
According to the Equation $$v^{2}-u^{2}=2gh$$,Where v is final Velocity, u is initial velocity, g is acceleration due to gravity and h is the height of the cliff
As 2gh and u is constant, the final velocity(v) is constant irrespective of the direction.
When a short bar magnet allowed to fall along the axis of horizontal metallic ring, then it will fall under the influence of gravity.
$$ \because \,s=ut+\dfrac{1}{2}g{{t}^{2}} $$
$$ here,\,\,u=0\,\,and\,\,t=\,1\,\,\sec $$
$$ s=\dfrac{1}{2}\times 10\times 1 $$
$$ =5\,\,m $$
The distance falls by the magnet in one second is less than $$5\,m$$ because when the magnet falls freely through the ring, flux increases and a field is induced whose direction is opposite to that of bar magnet. the magnet experiences a repulsion and hence acceleration of magnet becomes less than g.
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