Sunday, July 26, 2015

HC Verma solutions, Concepts of Physics, Part 1,Chapter 3, REST AND MOTION : KINEMATICS",'EXERCISES - Q1 to Q10

EXERCISES (Question number 1 to 10)

1. A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field. (a) What distance he has to walk to reach the field? (b) What is his displacement from his house to the field?          

Answer:   (a) Total distance traveled by the man 

                      = 50 m + 40 m + 20 m = 110 m.

               (b) His displacement is the vector joining points from his house to the field, which can be easily understood by the following figure:--

   Magnitude of displacement AB = √(30²+40²) = 50 m

Angle with horizontal is given by tan θ = 30/40 =3/4   

i.e. θ= tan-1  3/4, direction north to east. 



  



 2. A particle starts from the origin, goes along the X-axis to the point (20 m, 0) and then returns along the same line to the point (-20 m, 0). Find the distance and displacement of the particle during the trip. 

Answer:   Distance in forward motion = 20 m


Distance in backward motion = 20 m (up to origin) + 20 m (origin to further backward) = 40 m

Total Distance of the particle during the trip = 20 m + 40 m =60 m 


Displacement is the vector from the initial to the final position, its magnitude = 20 m due negative direction of the X-axis.



 


3. It is 260 km from Patna to Ranchi by air and 320 km by road. An airplane takes 30 minutes from Patna to Ranchi whereas the deluxe bus takes 8 hours. (a) Find the average speed of the plane. (b) Find the average speed of the bus. (c) Find the average velocity of the plane. (d) Find the average velocity of the bus.  


Answer:   (a) Average speed of the plane = 260 km/0.5 h =520 km/h  

(b) Average speed of the bus = 320 km/8 h = 40 km/h   

(c) Average velocity of the plane is a vector with magnitude = Shortest distance divided by the time taken = 260 km/0.5 h =520 km/h and direction Patna to Ranchi straight.

(d) The average velocity of the bus is a vector with magnitude = Shortest distance divided by the time taken = 260 km/8 h = 32.5 km/h and direction Patna to Ranchi straight. 

     



4. When a person leaves his home for sightseeing in his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. (a) What is the average speed of the car during this period? (b) what is the average velocity?


Answer:   Distance covered by car = 12416-12352 =64 km and time taken = 2 h 

(a) Average speed of the car = 64 km/ 2 h = 32 km/h  

(b) Displacement of the car is zero because it returns back to the same position, So average velocity =displacement/time =0/2 h= 0






5. An athlete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration?


Answer:  Difference of magnitudes of final and initial velocity = 18-0 = 18 km/h = 18000/3600 = 5 m/s,    Time = 2 s. 

 The magnitude of average acceleration =Difference of final and initial velocity magnitudes divided by time = 5÷2 = 2.5 m/s².  




6. The speed of a car as a function of time is shown in figure (3-E1). Find the distance traveled by car in 8 seconds and its acceleration.

The figure for question number 6.

Answer:  Distance traveled by car in 8 seconds is given by the area under the graph between 0 to 8 s. It is a triangle with a base of 8 s and a height of 20 m/s, Hence the area =½ x 20 m/s x 8 s = 80 m. 

Acceleration of the car is given by the change of velocity divided by time interval = (20-0)/8 =2.5 m/s². 


 Video Solution for Q - 6 is also here ↓



7. The acceleration of a cart started at t=0, varies with time as shown in figure (3-E2), Find the distance traveled in 30 seconds and draw the position-time graph.

The figure for question number 7.

Answer:  It is clear from the graph that the cart has a uniform acceleration of 5 ft/s² for the initial 10 s then uniform velocity for the next 10 s (because acceleration is zero) and then retardation in further next 10 s. The area between the graph and the time axis gives the change in velocity in the taken time interval.  Let us consider three parts of each 10 s.

    In the first 10 s change of velocity = 5 ft/s² x 10 s = 50 ft/s, Since the cart starts from rest at t=0, average velocity in this period = (50+0)/2 =25 ft/s, Distance traveled in this 10 s =25 ft/s x 10 s =250 ft. 

From t=10 s to 20 s area under the graph is zero, so the velocity change is zero, meaning that the velocity at t= 10 s (50 ft/s) remains unchanged. So distance traveled in this period = 50 ft/s x 10 s = 500 ft.
      From t=20 s to 30 s area under the graph = 5 ft/s² x 10 s = 50 ft/s but it is below the time axis, So this change of velocity is negative. This means that the cart moving with a velocity of 50 ft/s at t = 20 s comes to rest at t = 30 s. So average velocity in this period = 25 ft/s and distance travelled = 25 ft/s x 10 s = 250 ft.
 
     So the distance travelled = 250 ft + 500 ft+ 250 ft = 1000 ft.
The position-time graph may be drawn as follows,
Position-Time graph
     


8. Figure (3-E3) shows the graph of velocity versus time for a particle going along the X-axis. Find (a) the acceleration, (b) the distance traveled in 0 to 10 s and (c) the displacement in 0 to 10 s.

 
The figure for question number 8.

Answer:  Velocity at t=0 s is 2 m/s, velocity at t=10 s is 8 m/s

(a) Acceleration= Change in velocity/time interval = (8-2)/10 = 0.6 m/s²
(b) Distance traveled is given by the area under the graph between t=0 and t=10 s 
=½(2+8) x 10 = 50 m  
(c) Since the particle moves in one direction, the magnitude of displacement will be equal to the distance traveled = 50 m, along the X-axis.    





 

9. Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find (a) the average velocity during 0 to 10 s, (b) instantaneous velocity at 2, 5, 8 and 12 s.

The figure for question number 9.

 Answer:  (a) Displacement during 0 to 10 s = x = 100 m,   

Average velocity during 0 to 10 s = 100 m/ 10 s = 10 m/s

(b) Instantaneous velocity at n second,  vn is given by the slope of the graph at that instant  

v2 = slope of st line at 2 s = 50/2.5 = 20 m/s 

v5 = slope of st line at 5 s = zero   

v8 = slope of st line at 8 s = (100-50)/(10-7.5) =50/2.5 =20 m/s 

v12 = slope of st line at 12 s = -100/5 =-20 m/s  (because slope is negative) 




 


10. From the velocity-time plot shown in figure (3-E5), Find the distance traveled by the particle during the first 40 seconds. Also, find the average velocity during the period. 

 
The figure for question number 10
      Answer:  Distance traveled by the particle is given by the area between graph and X-axis, the area above X-axis is the distance traveled in the positive direction while the area below X-axis is the distance traveled in the negative direction. 

Distance in positive direction = area between t=0 s to t=20 s.

=½ x 20 x 5 = 50 m 

Distance in negative direction = area between t=20 s to t=40 s.

=½ x 20 x 5 = 50 m, So the total distance travelled = 50 m+ 50 m =100 m.  

Since the distance traveled in the positive and negative direction is equal, the particle returns to its original position, so the displacement is zero. 

Hence, the average velocity is zero. 

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(Next set of 10 solutions Q no 11 to 20 in blog link below)
NEXT SET OF SOLUTIONS from Q No 11 to 20

===<<<O>>>===

My Channel on YouTube  →  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com 





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

Click here for → Questions for Short Answer

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".

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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)

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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


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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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