Step-by-step explanation:
[tex]1)\:v = 2 {e}^{3t} + 5 {e}^{ - 3t} \\ differentiating \: w.r.t.t \: on \: both \: sides \\ acceleration =\\ \frac{dv}{dt} = \frac{1}{dt} (2 {e}^{3t} + 5 {e}^{ - 3t} ) \\ = 2 \times \frac{1}{dt} {e}^{3t} + 5 \times \frac{1}{dt} {e}^{ - 3t} \\ \\ = 2 \times {e}^{3t} \times 3 + 5 \times {e}^{ - 3t} \times ( - 3) \\ = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \frac{dv}{dt} = 6{e}^{3t} - 15{e}^{ - 3t} \\ \therefore \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3 \times 1} - 15 {e}^ { - 3 \times 1} \\ \bigg(\frac{dv}{dt} \bigg) _{t=1} = 6 {e}^{3} - 15 {e}^ { - 3 } \\ acceleration = \\ \purple{ \boxed{ \bold{\bigg(\frac{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 \times t \\ \therefore \: s= (2 {e}^{3t} + 5 {e}^{ - 3t}) \times t \\ \therefore \: (s)_{t=2} = (2 {e}^{3 \times 2} + 5 {e}^{ - 3 \times 2}) \times 2 \\ \therefore \: (s)_{t=2} = (2 {e}^{6} + 5 {e}^{ - 6}) \times 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}}}\\ [/tex]
