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Suggest a suitable physical situation for each of the following graphs.




(a) The given position-time graph shows that initially a body was at rest. Gradually, its velocity increases with time and attains an instantaneous constant value. Then, the velocity reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value.

A physical situation analogous to this graph arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime. 


(b) In the given velocity-time graph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.


(c) The given acceleration-time graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail. 
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Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, - 1.2 s.


At t= 0.3 sec,

Negative, Negative, Positive.

At t= 1.2 sec,

Positive, Positive, Negative

At t= -1.2 sec

Negative, Positive, Positive

For simple harmonic motion (SHM) of a particle, acceleration (a) is given by, 

a = – ω2x ;  ω → angular frequency       … (i) 

t = 0.3 s 

In this time interval, x is negative. Thus, the slope of the position-time graph will also be negative. Therefore, both position and velocity are negative.

However, using equation (i), acceleration of the particle will be positive.

t = 1.2 s 

In this time interval, x is positive. Thus, the slope of the x-graph will also be positive. Therefore, both position and velocity are positive.

However, using equation (i), acceleration of the particle comes to be negative. 

t = – 1.2 s 

In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive.

From equation (i), it can be inferred that the acceleration of the particle will be positive.

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Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.


The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time.

It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed is greatest in interval 3 and least in interval 2. 

For intervals 1 and 2, average velocity is positive as the slope is positive. For interval 3 slope is negative therefore, the average velocity is negative. 


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A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s-1. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?


Given comma space

Initial space velocity space of space the space ball comma space straight u space equals space 49 space straight m divided by straight s
Acceleration comma space straight a space equals space – space straight g space equals space – space 9.8 space straight m divided by straight s squared space

Case space 1 colon space

The space boy space throws space the space ball space when space the space lift space is space stationary. space
Considering space that space the space motion space of space the space ball space is space upwards comma space
final space velcoity space is space zero space at space the space highest space point. space

Now comma space using space first space equation space of space motion comma space

Time space of space ascent space is space given space by comma space

space space space space space space straight v space equals space straight u plus at space

rightwards double arrow space straight t space equals space fraction numerator left parenthesis straight v minus straight u right parenthesis over denominator straight a end fraction space

space space space space space space space equals space fraction numerator negative 49 over denominator negative 9.8 end fraction space equals space 5 space sec

Time space of space ascent space equals space time space of space descent. space

Therefore comma space

Total space time space taken space for space the space ball space to space return space
to space the space boy apostrophe straight s space hands space equals space 5 space plus space 5 space equals space 10 space sec. 

Case 2: 

Speed space with space which space the space lift space is space moving space up space equals space 5 space straight m divided by straight s

So comma space

Relatiove space velocity space of space the space ball space is space the space same
space as space it apostrophe straight s space initial space velocity.

straight i. straight e. comma space straight v space equals space 49 space straight m divided by straight s

Implies comma space the space ball space will space return space back space to space the space boy ’ straight s space
hand space after space 10 space straight s.
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A three-wheeler starts from rest, accelerates uniformly with 1 m s-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ?

Distance space covered space by space straight a space body space in space straight n to the power of th
second space is space given space by space the space relation comma

straight S subscript straight n space end subscript equals space straight u space plus space straight a fraction numerator space left parenthesis 2 straight n space minus space 1 right parenthesis over denominator 2 end fraction space space space space space space space space space... left parenthesis straight i right parenthesis space

where comma space

straight u space equals space Initial space velocity comma
straight a space equals space acceleration comma space
straight n space equals space time space intervals space equals space 1 comma 2 comma 3 comma... straight n

Here space we space have comma space

straight u space equals space 0 space and space straight a space equals space 1 space straight m divided by straight s squared

Therefore comma space

straight S subscript straight n space equals fraction numerator space left parenthesis 2 straight n space minus space 1 right parenthesis space over denominator 2 end fraction space space space space space space space space space space space space space space space space space space space... space left parenthesis ii right parenthesis space

This space relation space shows comma space

straight S subscript straight n space proportional to space straight n space space space space space space space space space space space space space space space... space left parenthesis iii right parenthesis space

Now, putting different values of n in equation (3), we have

n 1 2 3 4 5 6 7 8 9 10
Sn 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5
 
The graph between Sn and n will be a straight line as shown in the figure below. 



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