The speed-time graph of a particle movin | Class 11 Physics Chapter Motion in a straight Line, Motion in a straight Line NCERT Solutions

Welcome to the NCERT Solutions for Class 11 Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Exercise 1, Question 27: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 27:

The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

(Fig. 3.28)

What is the average speed of the particle over the intervals in (a) and (b)?

Answer:

(a) Distance travelled by the particle = Area under the given graph

= ½ x (10-0) x (12-0)  = 60 m

Average speed =  Distance / Time  = 60/10 = 6 m/s

 

(b) Let s1 and s2 be the distances covered by the particle between time

t = 2 s to 5 s and t = 5 s to 6 s respectively.

Total distance (s) covered by the particle in time t = 2 s to 6 s

s = s1 + s2 … (i)

For distance s1:

Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.

Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:

at

Where,

v = Final velocity of the particle

12 = 0 + a′ × 5

a′  = 12/5 = 2.4 m/s2

Again, from first equation of motion, we have

at

= 0 + 2.4 × 2 = 4.8 m/s

Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

su′ t + 1/2 a′t2

     =  4.8x3 + 1/2 x 2.4 x (3)2

     = 25.2m             ....(ii)

 

For distance s2:

Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.

From first equation of motion,

at (where v = 0 as the particle finally comes to rest)

0 = 12 + a″ × 5

a″  =  -12/5

    =  -2.4 m/s2

Distance travelled by the particle in 1s (i.e., between = 5 s and t = 6 s)

s u′' t + 1/2 a'′t2

  = 12 x a + ½ (-2.4) x (1)2

  = 12 - 1.2  = 10.8 m      ....... (iii)

From equations (i), (ii), and (iii), we get

s = 25.2 + 10.8 = 36 m

∴ Average speed  = 36/4 = 9 m/s


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Motion in a straight Line.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 27: The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the ....