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Question
An object was launched from a place P in constant speed to hit a target. At the 15th second, it was 1400 m from the target, and at the 18th second 800 m away. Find the distance between the place and the target
Solution
Let us take the time T along the x-axis and the distance D along the y-axis.
Given when time T = 15s, the distance D = 1400 m
The corresponding point is (15, 1400)
Also when time T = 18s, the distance D = 800 m.
The corresponding point is (18, 800)
The distance between the place and the target:
∴ The relation connecting T and D is the equation of the straight line joining the points (15, 1400) and (18, 800)
∴ `("T" - 15)/(18 - 15) = ("D" - 1400)/(800 - 1400)`
`("T" - 15)/3 = ("D" - 1400)/(- 600)`
T – 15 = `(1400 - "D")/200`
T = `(1400 - "D")/200 + 15` ......(1)
To find the distance between the target and the place,
Put T = 0 in equation (1)
(1) ⇒ 0 = `(1400 - "D")/200 + 15`
⇒ 0 = `(1400 - "D" + 300)/200`
4400 – D = 0
⇒ D = 4400 m.
Required distance = 4400 m.
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