Question

Help need full solution i &ii​

Answer 1

Explanation:

`1)\:v = 2 {e}^(3t) + 5 {e}^( - 3t) \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ (dv)/(dt) = (1)/(dt) (2 {e}^(3t) + 5 {e}^( - 3t) ) \\ = 2 * (1)/(dt) {e}^(3t) + 5 * (1)/(dt) {e}^( - 3t) \\ \\ = 2 * {e}^(3t) * 3 + 5 * {e}^( - 3t) * ( - 3) \\ = 6{e}^(3t) - 15{e}^( - 3t) \\ \therefore (dv)/(dt) = 6{e}^(3t) - 15{e}^( - 3t) \\ \therefore \bigg((dv)/(dt) \bigg) _(t=1) = 6 {e}^(3 * 1) - 15 {e}^ { - 3 * 1} \\ \bigg((dv)/(dt) \bigg) _(t=1) = 6 {e}^(3) - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg((dv)/(dt) \bigg) _(t=1) = \bigg(\frac{6 {e}^(6) - 15}{ {e}^(3) } \bigg) \: m {s}^( - 2) }}} \\ \\ 2) \: let \:s \: be \: the \: total \: distance \: travelled \\ \therefore \: s = v * t \\ \therefore \: s= (2 {e}^(3t) + 5 {e}^( - 3t)) * t \\ \therefore \: (s)_(t=2) = (2 {e}^(3 * 2) + 5 {e}^( - 3 * 2)) * 2 \\ \therefore \: (s)_(t=2) = (2 {e}^(6) + 5 {e}^( - 6)) * 2 \\ \therefore \: (s)_(t=2) = 4 {e}^(6) + 10{e}^( - 6) \\ \red{ \boxed{ \bold{\therefore \: (s)_(t=2) = \bigg(\frac{4 {e}^(12) + 10}{{e}^( 6)} \: \bigg)m}}}\\`