Question

A body moving along straight line covers a distance
in three equal parts with speed 2 m/s, 3 m/s and
4 m/s, then its average speed is
(1) 12 m/s
(2)13/12 m/s
(3) 36/13 m/s
(4)12/13 m/s​

Answer 1

Answer:

use formula 3v1v2v3/v1v2+v2v3+v3v1

Explanation:

v1=2,V2=3,V3=4

average velocity = 3×2×3×4/2×3+3×4×4×2

= 72/26

=36/13m/s

Answer 2

The average speed is (3) 36/13 m/s

Given - Different speed

Find - Average speed

Solution - Let the three equal distance be d each. Average speed = total distance/total time

Total distance = d + d + d

T1 = d/2

T2 = d/3

T3 = d/4

Total time = d/2 + d/3 + d/4

Taking LCM, we get 12

Total time = $\frac{(d \times 6) + (d \times 4) + (d \times 3)}{12}$

Total time = $\frac{6d + 4d + 3d}{12}$

Total time = 13d/12

Average speed = $\frac{3d}{ \frac{13d}{12} }$

Average speed = $\frac{3d \times 12}{13d}$

Average speed = $\frac{36d}{13d}$

Average speed = 36/13

Hence, the average speed is (3) 36/13 m/s.

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