If 𝑎 = 4𝑡^(− 3/2) , 𝑠 = 16 𝑤ℎ𝑒𝑛 𝑡 = 4, 𝑎𝑛𝑑 𝑠 = 25 𝑤ℎ𝑒𝑛 𝑡 = 6, find the equation of motion 𝑠 = 𝑓(𝑡) and the velocity function 𝑣(𝑡).
v(t)=∫a(t)dt=∫4t−3/2dt=−8t+C1v(t)=\int a(t)dt= \int 4t^{-3/2}dt= \frac{-8}{\sqrt{t}} +C_1v(t)=∫a(t)dt=∫4t−3/2dt=t−8+C1
s(t)=∫v(t)dt=∫(−8t−0.5+C1)dt=−16t+C1t+C2s(t)= \int v(t)dt =\int( -8t^{-0.5}+C_1)dt =-16\sqrt{t} +C_1t + C_2s(t)=∫v(t)dt=∫(−8t−0.5+C1)dt=−16t+C1t+C2
16=-32+4C1+C2
C2=48+4C1
25=−166+6C1+48+4C125=-16 \sqrt{6}+ 6C1+48+4C125=−166+6C1+48+4C1
10C1=−23+16610C_1=-23+16 \sqrt{6}10C1=−23+166
C1=1.66−2.3C_1=1.6 \sqrt{6}-2.3C1=1.66−2.3
C2=38.8+6.46C_2=38.8+6.4 \sqrt{6}C2=38.8+6.46
Answer: v(t)=−8t+1.66−2.3v(t)= \frac{-8}{\sqrt{t}}+1.6\sqrt{6}-2.3v(t)=t−8+1.66−2.3
s(t)=−16t+t(1.66−2.3)+38.8+6.46s(t)=-16\sqrt{t}+t(1.6 \sqrt{6}-2.3) +38.8+6.4\sqrt{6}s(t)=−16t+t(1.66−2.3)+38.8+6.46
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