Solution: Assume $ V(t) = at^2 + bt + c $. From $ V(1) = a + b + c = 120 $, $ V(2) = 4a + 2b + c = 200 $, $ V(3) = 9a + 3b + c = 300 $. Subtract first equation from the second: $ 3a + b = 80 $. Subtract second from the third: $ 5a + b = 100 $. Subtract these: $ 2a = 20 $ â†’ $ a = 10 $. Then $ 3(10) + b = 80 $ â†’ $ b = 50 $. From $ a + b + c = 120 $: $ 10 + 50 + c = 120 $ â†’ $ c = 60 $. Thus, $ V(t) = 10t^2 + 50t + 60 $. For $ t = 4 $: $ V(4) = 10(16) + 50(4) + 60 = 160 + 200 + 60 = 420 $. Final answer: $ oxed{420} $.

Sun, 03 May 2026 05:20:27 +0000

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