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Page No 60: - Chapter 3 Motion In A Straight Line class 11 ncert solutions Physics - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Question 3.27:

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) = 0 s to 10 s, (b) = 2 s to 6 s.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_m391b9f1b.jpg

(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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_m1f18ffe5.gif

Average speed = https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_m2d30319b.gif

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

= 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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_m7d8b021c.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_78a39c72.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_m13bd2e1f.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_692f4bc4.gif

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

s = 25.2 + 10.8 = 36 m

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4375/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_71b66cf8.gif

Question 3.28:

The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4376/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_301e2846.jpg

Which of the following formulae are correct for describing the motion of the particle over the time-interval t2 to t1?

(a) x(t2) = x(t1) + v(t1)(t2–t1) + (https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4376/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_eeecab0.gif )a(t2–t1)2

(b) v(t2)= v(t1+ a(t2–t1)

(c) vAverage (x(t2– x(t1)) / (t2 – t1)

(d) aAverage (v(t2) – v(t1)) / (t2 – t1)

(e) x(t2= x(t1) + vAverage(t2 – t1) + (https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/181/4376/NS_6-10-08_Ravinder_11_Physics_3_28_NRJ_html_eeecab0.gif )aAverage(t2 – t1)2

(f) x(t2) – x(t1) = area under the v–t curve bounded by the t-axis and the dotted line shown.

Answer:

The correct formulae describing the motion of the particle are (c), (d) and, (f)

The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.

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