Question

x=5t^3 + 6t^2 calculate velocity at t =1s​



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Answer 1

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{x = {5t}^{3} + {6t}^{2}}$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}}$

$\:\:\:\:\bullet\:\:\:\sf{ Velocity \: at \: (t = 1)sec}$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}}$

☯ Finding velocity at (t = 1) sec

$\dashrightarrow\:\: \sf{ x = {5t}^{3} + {6t}^{2} }$

$\dashrightarrow\:\: \sf{ \dfrac{dx}{dt} = {5t}^{3} + 6 {t}^{2} }$

$\dashrightarrow\:\: \sf{ v = 5 \times 3 \times {t}^{3 - 1} + 6 \times 2 \times t^{2 - 1} }$

$\dashrightarrow\:\: \sf{ v = 15 {t}^{2} + 12t}$

$\dashrightarrow\:\: \sf{v = 15 \times {1}^{2} + 12 \times 1}$

$\dashrightarrow\:\:{\boxed{\sf{\purple{v = 27\:m/s }}}}$

☢ $\underline{\boxed{\sf{{\red{Velocity}} \:{\orange{ of }}\: {\blue{particle\: at }}\:{\purple{ (t = 1) \:is}} \: {\pink{27 \: m/s}}}}}$