Question #193663

 A deer is moving with its position varying as a function of time according to the equation: x = 18 + 1.75t – 0.04t 2 Where x is in cm and t is in seconds.

i. What is the deer’s: - initial position? - initial velocity? - initial acceleration?

ii. Determine the time when the deer’s velocity is zero?

iii. How long after starting does the deer’s displacement become zero? Where is the deer then?


1
Expert's answer
2021-05-16T17:59:11-0400

i. x0=18 (cm)x_0=18\ (cm) ; v0=x(0)=1.750.080=1.75 (cm/c)v_0=x'(0)=1.75-0.08\cdot0=1.75\ (cm/c)

a0=v(0)=0.08 (cm/c2)a_0=v'(0)=-0.08\ (cm/c^2)


ii. v=1.750.08t0=1.750.08tt=21.875 (s)v=1.75-0.08t\to 0=1.75-0.08t\to t=21.875\ (s)


iii. Δx=0\Delta x=0 if x=18 (cm)x=18\ (cm) . So,


18=18+1.75t0.04t2t=43.75 (s)18=18+1.75t-0.04t^2\to t=43.75\ (s)

The deer is at the point x=18 (cm)x = 18\ (cm) .






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