Question

If x=12t^3_4t^2+6t+8, find the value of velocity at t=4second

Answer 1

Answer:

• x = 12t³ - 4t² + 6t + 8

Instantaneous velocity:

$\longrightarrow\sf v=\dfrac{dx}{dt}$

$\longrightarrow\sf v=\dfrac{d}{dt}( {12t}^{3} - 4 {t}^{2} + 6t + 8)$

$\longrightarrow\sf v=3 \times 12 {t}^{2} - 2 \times 4 t + 6 + 0$

$\longrightarrow\sf v=36 {t}^{2} - 8t + 6$

At time t = 4 second velocity becomes:

$\longrightarrow\sf v=36 {(4)}^{2} - 8(4)+ 6$

$\longrightarrow\sf v=36 \times 16 - 32+ 6$

$\longrightarrow\sf v=576 - 32+ 6$

$\longrightarrow\sf v=576 - 32+ 6$

$\longrightarrow\sf v=544+ 6$

$\longrightarrow\bf v=550 \: {ms}^{ - 1}$

Answer 2

$\\ \maltese \: \large \underline \orange{\pmb {\sf Correct \: Question:- }}$

• if $\sf x = 12t³-4t²+6t+8$ then find the value of velocity at t = 4 seconds.

$\\ \maltese \: \large \underline \orange{\pmb {\sf Solution:- }}$

• We know that :-

$\leadsto \large\underline{ \boxed {\pink{{ \sf{velocity = \frac{dx}{dt} } }}}} \bigstar$

$: \implies \sf velocity = \dfrac{d}{dt} ( {12t}^{3} - {4t}^{2} + 6t + 8 )$

$: \implies \sf velocity = (3 \times 12 {t}^{2}) - (2 \times 4t) + 6 + 0$

$: \implies \boxed {\pink{{ \sf velocity = 36 {t}^{2} - 8t + 6}}} \bigstar$

$\\ \dag \: \underline{ \frak{substituting \: given \: value \: of \: time \: in \: above \: equation : - }}$

$: \implies \sf velocity = 36( {4})^{2} - 8(4) + 6$

$: \implies \sf velocity = 36 \times 16 - 32 + 6$

$: \implies \sf velocity = 576 - 32 + 6$

$: \implies \boxed{ \purple{{\sf velocity = 550 \: m {s}^{ - 1} } }}\bigstar$

$\: \: \: \: \: \: \: \: \: \\\therefore \underline{ \sf 550 \: m {s}^{ - 1} \: is \: the \: required \: answer \: }$

Hope it's helpful :D