Example 5 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at May 29, 2018 by Teachoo
Examples
Example 1 (a)
Example 1 (b)
Example 1 (c) Important
Example 1 (d)
Example 2 Important
Example 3 Important
Example 4
Example 5 You are here
Example 6 Important
Example 7
Example 8
Example 9 (i)
Example 9 (ii) Important
Example 10
Example 11
Example 12
Example 13 Important
Example 14 Important
Example 15 Important
Example 16
Example 17
Example 18 Important
Example 19
Example 20
Example 21 Important
Example 22 Important
Example 23
Example 24 Important
Example 25 Important
Chapter 10 Class 11 Straight Lines (Term 1)
Serial order wise
- Ex 10.1
- Ex 10.2
- Ex 10.3
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Examples
- Miscellaneous
Transcript
Example 5 In Fig 10.9, time and distance graph of a linear motion is given. Two positions of time and distance are recorded as, when T = 0, D = 2 and when T = 3, D = 8. Using the concept of slope, find law of motion, i.e., how distance depends upon time. Let A = (0, 2) , B = (3, 8) , C = (T, D) Points A,B, C lie on the line So, A, B & C are collinear Slope of AB = Slope of BC We know that slope of a line through the points (x1, y1)(x2, y2)is m = ( 2 1)/( 2 1) D 8 = 2T 6 D = 2T 6 + 8 D = 2T + 2 D = 2(T + 1) Here, we can see that as T increases , D increases Hence, Distance depends on time.
