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Katena32 [7]
3 years ago
15

The position-time equation for a cheetah chasing an antelope is:

Physics
1 answer:
pochemuha3 years ago
4 0

Answer:

x = 1.6 + 1.7 t^2      omitting signs

a) at t = 0     x = 1.6 m

b) V = d x / d t = 3.4 t

at t = 0     V = 0

c) A = d^2 x / d t^2 = 3.4     (at t = 0  A = 3.4 m/s^2)

d)  x = 1.6 + 1.7 * (4.4)^2 = 34.5    (position at 4.4 sec = 34.5 m)

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Answer:

a) t=24s

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x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

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w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

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Given: m=1.2 kg

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a) A(t)=A₀/8

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⇒t=\frac{2m*ln(8)}{b}

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t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

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T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

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let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

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