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

Welcome to the NCERT Solutions for Class 11th Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Exercise 1, Question 27: the speed time graph of a particle moving along a....
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

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