Question

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

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Velocity=27\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies x = {5t}^{3} + {6t}^{2} \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Velocity \: at \: (t = 1)sec =?$

• According to given question :

$\bold{As \: we \:know \: that} \\ \tt: \implies x = {5t}^{3} + {6t}^{2} \\ \\ \tt: \implies \frac{dx}{dt} = {5t}^{3} + 6 {t}^{2} \\ \\ \tt: \implies v = 5 \times 3 \times {t}^{3 - 1} + 6 \times 2 \times t^{2 - 1} \\ \\ \tt: \implies v = 15 {t}^{2} + 12t \\ \\ \tt: \implies v = 15 \times {1}^{2} + 12 \times 1 \\ \\ \green{\tt: \implies v = 27 {m/s}} \\ \\ \green{\tt \therefore Velocity \: of \: particle \: at \: (t = 1) \:is \: 27 \: m/s}$