Answer with Explanation:
We are given that
[tex]y=(18-2x^2) km[/tex]
[tex]x=(4t-3)km[/tex]
Differentiate x and y w.r.t t
[tex]\frac{dx}{dt}=4[/tex]
[tex]\frac{dy}{dt}=-4x\frac{dx}{dt}[/tex]
[tex]\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)[/tex]
[tex]v_x=\frac{dx}{dt}=4[/tex]
[tex]v_y=\frac{dy}{dt}=-16(4t-3)[/tex]
Substitute t=1
[tex]v_x=4[/tex]
[tex]v_y=-16(4-3)=-16[/tex]
Magnitude of velocity=[tex]\mid v\mid=\sqrt{v^2_x+v^2_y}[/tex]
[tex]\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s[/tex]
Hence, the magnitude of the missile's velocity=16.49 m/s
[tex]a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0[/tex]
[tex]a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64[/tex]
Substitute t=1
[tex]a_x=0,a_y=-64[/tex]
[tex]\mid a\mid=\sqrt{a^2_x+a^2_y}[/tex]
[tex]\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2[/tex]
Hence, the magnitude of acceleration when t=1 s=[tex]64m/s^2[/tex]
