Question

In the equation x = t3+3t2+4t+5 find velocity at t = 2 sec and t = 3 sec​

Answer 1

$\huge\sf{Question :-}$

In the equation x = t3+3t2+4t+5 find velocity at t = 2 sec and t = 3 sec.

$\huge \sf{Answer :-}$

given equation

${t}^{3} +3 {t}^{2} + 4t + 5$

$v = \frac{dx}{dt}$

$\frac{dx}{dt} ( {t}^{3} + 3 {t}^{2} + 4t + 5)$

$\: \: \: = 3 {t}^{2} + 6t + 4$

for t = 2 sec

$v = 3 {t}^{2} + 6t + 4$

$\: \: \: \: = 3(2 \times 2) + 6(2) + 4$

$\: \: \: \: = 12 + 12 + 4$

=28m/sec

for t = 3 sec

$v = 3 {t}^{2} + 6t + 4$

$\: \: \: \: = 3(3 \times 3) + 6(3) + 4$

$\: \: \: \: = 27 + 18 + 4$

= 49 m/sec