Question

An object was launched from a place P in constant speed to hit a target. At the 15^th second, it was 1400 m from the target, and at the 18^th second 800 m away. Find the distance between the place and the target

Answer 1

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.