Answer:
The distance covered is 113.75 m
Step-by-step explanation:
As per the question:
The initial velocity of the train, v = 20 m/s
The final velocity of the train, v' = 6 m/s
Uniform deceleration, a = 1.6
Or uniform acceleration, a = - 1.6
<em>Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.</em>
Now, to find the distance covered, s:
Using the eqn of Kinemetics:
(a) The vertices after the reflection in the line x = 5 are F'(4,6) , G'(2,8) , H'(2,3) .
and (b) after translation , the vertices are F''(-1,-3) , G'' (1,-1) , H''(1,-6) .
In the question a triangle FGH with vertices F(6,6) , G(8,8) and H(8,3) is given
Part (a)
the rule for reflection of point (x,y) by the line x = p is
(x,y) → (2×p - x , y)
reflection by the line x = 5 ,
we get
the vertices as
(2×5-6,6) = (4,6)
G'(2×5-8,8) = (2,8)
H'(2×5 - 8 ,3) = (2,3)
Part (b)
the rule of translation is given as (x,y)→ (x - 7, y - 9)
So , the vertices after translation are
F''(6-7 , 6-9) = (-1,-3)
G''(8-7,8-9) = (1,-1)
H''(8-7,3-9) = (1,-6)
Therefore , (a) The vertices after the reflection in the line x = 5 are F'(4,6)
G'(2,8) , H'(2,3) .
and (b) after translation , the vertices are F''(-1,-3) , G'' (1,-1) , H''(1,-6) .
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