Problem 1 :
The normal boiling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F.
(i) Find the linear relationship between C and F
(ii) the value of C for 98.6°F and
(iii) the value of F for 38°C
Solution :
(i) To find the linear relationship between C and F, let us write them as points
(100, 212) and (0, 32)
(y - 212)/(32 - 212) = (x - 100)/(0 - 100)
(y - 212)/(-180) = (x - 100)/(- 100)
10(y - 212) = 18(x - 100)
5(y - 212) = 9(x - 100)
y - 212 = (9/5) (x - 100)
y - 212 = (9/5) (x) - 100(9/5)
y - 212 = (9/5) (x) - 180
y = (9/5) (x) - 180 + 212
y = (9/5) (x) + 32
By replacing C and F by x and y respectively, we get
F = (9/5) (C) + 32
(ii) the value of C for 98.6^{◦}F and
If F = 98.6 then C = ?
98.6 = (9/5)C + 32
(98.6 - 32) ⋅ (5/9) = C
C = 37°
(iii) the value of F for 38°C
If C = 38 then F = ?
F = (9/5)C + 32
F = (9/5) ⋅ (38) + 32
= 100.4°
Problem 2 :
An object was launched from a place P in constant speed to hit a target. At the 15th second it was 1400m away from the target and at the 18th second 800m away. Find
(i) the distance between the place and the target
(ii) the distance covered by it in 15 seconds.
(iii) time taken to hit the target.
Solution :
Let us write time and distance in a point form
(15, 1400) and (18, 800)
Let us find the equation of line to represent the linear relation ship between the time and distance.
(y - 1400)/(800 - 1400) = (x - 15)/(18 - 15)
(y - 1400)/(-600) = (x - 15)/3
y - 1400 = -200(x - 15)
y = -200x + 3000 + 1400
y = -200x + 4400
By replacing x and y by t and D respectively, we get
D = -200t + 4400
(i) the distance between the place and the target
To find the distance between place and target, we have to apply 0 instead of t.
y = -200(0) + 4400
y = 4400 m
(ii) the distance covered by it in 15 seconds.
D = -200t + 4400
t = 15
D = -200(15) + 4400
D = -3000 + 4400
D = 1400
(iii) time taken to hit the target.
When we achieve the target the distance will become 0.
So, D = 0
0 = -200t + 4400
200 t = 4400
t = 4400/200 = 22 seconds
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