Respuesta :
initial velocity is given as 41 km/h at 60 degree North of West
[tex]v_i = -41 cos60\hat i + 41 sin60 \hat j[/tex]
[tex]v_i = -20.5 \hat i + 35.5 \hat j[/tex]
After some time the velocity is given as
[tex]v_f = 25 \hat j[/tex]
now we can find the acceleration
[tex]a = \frac{v_f - v_i}{t}[/tex]
[tex]a = \frac{25\hat j - (-20.5 \hat i + 35.5 \hat j)}{3}[/tex]
[tex]a = 6.83 \hat i - 3.5 \hat j[/tex]
now the distance is given by
[tex] d = v*t + \frac{1}{2} at^2[/tex]
[tex]d = (-20.5 \hat i + 35.5 \hat j)*4.5 + 0.5*(6.83 \hat i - 1.5 \hat j)* 4.5^2[/tex]
[tex]d = -92.25\hat i + 159.75\hat j + 69.15\hat i - 15.19\hat j[/tex]
[tex]d = -23.1 \hat i + 144.56\hat j[/tex]
so the magnitude of distance is
[tex]d = \sqrt{23.1^2 + 144.56^2} = 146.4 km[/tex]
Explanation:
It is given that, the eye of a hurricane passes over grand Bahama island in a direction 60° north of west with a speed of 41.0 km/h. Since, velocity is a vector quantity. It can be written as :
[tex]v_i=41cos(60)i+41sin(60)j[/tex]
[tex]v_i=(20.5i+35.5j)\ km/h[/tex]
Three hours later the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h, [tex]v_f=25j\ km/h[/tex]
We have to find how far from grand Bahama is the hurricane 4.50 h after it passes over the island.
[tex]s=ut+\dfrac{1}{2}at^2[/tex].............(1)
Acceleration, [tex]a=\dfrac{v_f-v_i}{t}[/tex]
[tex]a=\dfrac{25j-(20.5i+35.5j)}{3}[/tex]
[tex]a=\dfrac{(-20.5i+10.5j)}{3}[/tex]
a =( -6.8i+3.5j ) km/h²
Put the value of a in equation (1) and calculating x and y component of displacement as :
[tex]s_x=v_i_xt+\dfrac{1}{2}a_xt^2[/tex]
[tex]s_x=20.5\times 4.5+\dfrac{1}{2}\times (-6.8)\times 4.5^2[/tex]
[tex]s_x=23.4\ km[/tex]
and
[tex]s_y=v_i_yt+\dfrac{1}{2}a_yt^2[/tex]
[tex]s_y=35.5\times 4.5+\dfrac{1}{2}\times (3.5)\times 4.5^2[/tex]
[tex]s_y=195.18\ km[/tex]
Magnitude of distance, [tex]s=\sqrt{s_x+s_y}[/tex]
[tex]s=\sqrt{23.4^2+195.18^2}[/tex]
s = 196.5 km
Hence, this is the required solution.
