(a) P(-1, 4), Q(2, 3), R(x, y) and S(-2, 3) are the verticles of a parallelogram. Find the value of x and y.
(b) A particle starts from rest and moves in a straight line. It attains a velocity of 20ms\(^{-1}\) after travelling a distance of 8 metres. Calculate;
(a) To find the value of x and y, we can use the fact that opposite sides of a parallelogram are parallel and have equal length.
From the given information, we know that PQ and SR are parallel, and PQ = SR since they are opposite sides of the parallelogram. Therefore, we can use the distance formula to find the length of PQ and SR:
PQ = sqrt((2 - (-1))^2 + (3 - 4)^2) = sqrt(3^2 + (-1)^2) = sqrt(10)
SR = sqrt((-2 - x)^2 + (3 - y)^2)
Since PQ = SR, we have:
sqrt(10) = sqrt((-2 - x)^2 + (3 - y)^2)
Squaring both sides, we get:
10 = (2 + x)^2 + (3 - y)^2
Expanding the squares, we get:
10 = 4 + 4x + x^2 + 9 - 6y + y^2
Simplifying, we get:
x^2 + y^2 + 4x - 6y - 3 = 0
We can rewrite this equation as:
(x + 2)^2 + (y - 3)^2 = 13
This is the equation of a circle with center (-2, 3) and radius sqrt(13). Therefore, any point (x, y) that lies on this circle will form a parallelogram with P(-1, 4), Q(2, 3), and S(-2, 3).
(b)
(i) We know that the initial velocity of the particle is zero, and that its final velocity is 20ms-1 after travelling a distance of 8 metres. We can use the formula:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance travelled.
Plugging in the given values, we get:
(20)^2 = 0^2 + 2a(8)
Simplifying, we get:
a = 100/8 = 12.5ms-2
Therefore, the acceleration of the particle is 12.5ms-2.
(ii) We can use the formula:
s = ut + 0.5at^2
where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time taken. Since the initial velocity is zero, the formula simplifies to:
s = 0.5at^2
Plugging in the given values, we get:
40 = 0.5(12.5)t^2
Simplifying, we get:
t^2 = 6.4
Taking the square root of both sides, we get:
t = 2.53 seconds (rounded to two decimal places)
Therefore, the time taken