Saturday, August 1, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 3, REST AND MOTION : KINEMATICS",'EXERCISES - Q11 to Q20

EXERCISES (Question number 11 to 20)

REST AND MOTION

11. Figure (3-E6) shows x - t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero.
The figure for Question number 11
 

Answer: The average velocity in t second will be zero if the displacement at time t will remain equal to the displacement at t=0 which is 20 m. So we draw a line parallel to the time axis from 20 m on the displacement axis to meet the curve at P (See the figure below). From P draw a line perpendicular to the time axis PQ. The time at Q gives the instant when the displacement is again 20 m and average velocity zero. From eye estimation, t=12 s at Q.
Answer figure for Q no 11




 

12. A particle starts from point A and travels along the solid curve shown in Figure (3-E7). Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.

The figure for Question number 12

Answer: Instantaneous velocity at B will be along the tangent at point B, while the direction of average velocity will be along the direction of displacement AB. So both to have the same direction we draw a line from point A to curve such that it is tangent to the curve. The point where it is tangent to the curve is the required point B. It has been shown in the following figure:--  
Answer for Question number 12

The direction of average velocity is AB and the direction of instantaneous velocity at B is BC, both in the same direction. From the figure, the approximate position of B is x= 5 m, y= 3 m.

  


13. An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance traveled during the period of acceleration. 


Answer: Initial velocity u=4.0 m/s² 

acceleration a=1.2 m/s². 

time t=5.0 s 

from the equation of accelerated motion on a straight line x=ut+½at²  

x=4x5+½x1.2x5² =20+15= 35 m 


 




14. A person traveling at 43.2 km/h applies the break giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?


Answer: We should convert the given data into matching units.

u =43.2 km/h =43.2 x 1000/3600 m/s =12 m/s. 
a = -6.0 m/s² , Final velocity v = 0 m/s.  

From the equation of accelerated motion on a straight line, 

v² = u² +2ax   

→ 0² = (12)²-2 *6*x   

→x = (12)²/12 = 12 m.



  

15. A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by train, (b) the maximum speed attained by the train and, (c) the position(s) of the train at half the maximum speed.


Answer: We solve it in two parts, 1. accelerated motion and 2. decelerated motion. 

For 1. accelerated motion 

u =0, a =2.0 m/s². t =½*60 =30 seconds, Now final velocity is given by v = u +at =0 +2*30 =60 m/s 

Since after this velocity brakes are applied, so it is the 

(b) maximum speed attained = 60 m/s.

Distance traveled x = ut +½at² = 0 +½*2*30² =900 m =0.9 km 

Let the distance traveled at half the maximum speed (30 m/s) be y, Now from the equation v²=u²+2ax,

30²=0²+2*2*y    → y=900/4 =225 m .........(1)

2. For decelerated motion  

u =60 m/s, v =0 m/s, t =60 s From the equation v =u +at 

a =(0-60)/60 = -1 m/s² 

Distance  traveled = ut +½at² =60*60 +½*-1*60² 

                             = 3600-1800 =1800 m =1.8 km 

Hence answer for (a) The total distance moved = 0.9 km + 1.8 km 

=2.7 km.   

Let the distance moved at half the speed in this part be z  

From v² =u² +2ax 

→ 30² =60² +2*-1*z 

→ 2z =60² -30²=2700 m  

→ z =1350 m =1.35 km ....(2)

(c) There will be two positions at half the maximum speed (From the point of start),

      From ..(1) =225 m and From ...(2) 1.35 km+0.9 km = 2.25 km  


 




16. A bullet traveling with a velocity of 16 m/s penetrates a tree trunk and comes to rest 0.4 m. Find the time taken during the retardation.

Answer: Initial velocity u = 16 m/s,  distance moved, x =0.4 m,  Final velocity, v =0.

 From the kinetics equation of motion v² =u² +2ax

0 =16² +2*a*0.4  

→ a =-16²/0.8  =-320 m/s² (-ve sign is for the retardation)

Time taken during the retardation is given by the equation v= u +at 
→ 0= 16-320*t   → t=16/320 =1/20 =0.05 s.
 


  



17. A bullet going with a speed of 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find the deceleration.  

Answer: First make the given units uniform.

Traveled distance x =5.0 cm =0.05 m. 

Initial velocity, u =350 m/s, Final velocity, v =0.

Acceleration =?

Now from the kinematics equation of accelerated motion,

 v² = u² + 2ax  

→ 0² =350² +2*a*0.05 

→ a = -350²/0.1 

        = -1225000 m/s² (-ve sign is for retardation/deceleration) 

So deceleration = 12.2 x 105 m/s². 
 


 


18. A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed of 18 km/h, find (a) the average velocity during this period, and (b) the distance traveled by the particle during this period. 


Answer: Given u =0, t =5.0 s,  v =18 km/h =18000/3600 m/s =5 m/s

(a) Average velocity =(u +v)/2 = (0 +5)/2 m/s =2.5 m/s.
(b) Distance traveled in this period = Av. Velocity*Time
                                                     = 2.5 m/s* 5.0 s =12.5 m.

We can also verify the average velocity as the total distance traveled divided by the time i.e. = x/t. See below.
(a) From v = u +at 
→ 5 =0 +a*5 
→ a =1 m/s². 
From the equation, x =ut +½at² 
→ Average velocity x/t =u+½at = 0 +½*1*5 =2.5 m/s.
It is the same as calculated earlier. 

19. A driver takes 0.2 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s², find the distance traveled by the car after he sees the need to put the brakes on.    

Answer: u =54 km/h =54000/3600 =15 m/s. 

First, the distance traveled during reaction time = speed x time 

= 15 m/s x 0.2 s =3 m. 

Now for the decelerated motion, u =15 m/s, v =0 m/s. a = -6.0 m/s² 

From v² = u² +2ax 

→0² =15² +2*-6*x  

→x =225/12 m =18.75 m 

So, the total distance traveled by the car after the driver sees the need to put the brakes on = 18.75 +3 =21.75 m ≈22 m  




20. Complete the following table:-
 
Table for Question Number 20


Answer: Braking distance  =(u² -v²)/2a =u²/2a  

Distance in reaction time = ut 
Total stopping distance TSD=ut+u²/2a 

Now for u=54 km/h =15 m/s, t=0.2 s 
Braking distance a =18.75 m  ≈19 m 
TSD b =22 m
But for t =0.3 s   and u =72 km/h = 20m/s , 
Braking distance c = 20²/(2*6) =400/12 = 33.33 m ≈ 33 m and TSD d = 20*0.3+33 =39 m 
Now for deceleration 7.5 m/s², t=0.2 s, u=15m/s, 
Braking distance e  = 15²/(2*7.5) =225/15 = 15 m and 
TSD f = 15*0.2+15 =18 m   
Again for deceleration 7.5 m/s², t =0.3 s, u =20m/s,
 Braking distance g = 20²/(2*7.5) =400/15 = 26.67 m ≈27 m, 
and TSD h = 20*0.3+27 =33 m.   Now it can be filled in the given table as below:-
 
Answer for question number 20


--------------------------------------------------------------------------------  Next set of 10 Questions and Answers from 21 to 30  in the next blog  at the following link
 Question no. 21 to 30 
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CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion



EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)

CHAPTER- 11 - Gravitation



EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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