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madam [21]
3 years ago
6

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

, where t is in seconds, determine the magnitudes of the missile's velocity and acceleration when t=1 s.
Physics
1 answer:
muminat3 years ago
3 0

Answer with Explanation:

We are given that

y=(18-2x^2) km

x=(4t-3)km

Differentiate x and y w.r.t t

\frac{dx}{dt}=4

\frac{dy}{dt}=-4x\frac{dx}{dt}

\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)

v_x=\frac{dx}{dt}=4

v_y=\frac{dy}{dt}=-16(4t-3)

Substitute t=1

v_x=4

v_y=-16(4-3)=-16

Magnitude of velocity=\mid v\mid=\sqrt{v^2_x+v^2_y}

\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s

Hence, the magnitude of the missile's velocity=16.49 m/s

a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0

a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64

Substitute t=1

a_x=0,a_y=-64

\mid a\mid=\sqrt{a^2_x+a^2_y}

\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2

Hence, the magnitude of acceleration when t=1 s=64m/s^2

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Answer:

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