Question

The acceleration of a particle at any time t ≥ 0 is given by 𝑎 = 𝑑𝑣 𝑑𝑡 = 12 cos 2𝑡 𝑑𝑡𝑖 − 8 sin 2𝑡 𝑗 + 16 𝑡 𝑘 , if the velocity v and displacement r are zero at t=0, find 𝑣 𝑎𝑛𝑑 𝑟 at any time.

Answer 1

Let's find velocity:

$v(t) = \int {a(t)dt} = \int {12\cos 2t} dt\overrightarrow i - \int {8\sin 2t} dt\overrightarrow j + \int {16tdt} \overrightarrow k = \left( {6\sin 2t + {C_1}} \right)\overrightarrow i + \left( {4\cos 2t + {C_2}} \right)\overrightarrow j + \left( {8{t^2} + {C_3}} \right)\overrightarrow k$

By condition, v(0)=0. Then

$6\sin 0 + {C_1} = 0 \Rightarrow {C_1} = 0$

$4\cos 0 + {C_2} = 0 \Rightarrow 4 + {C_2} = 0 \Rightarrow {C_2} = - 4$

$0 + {C_3} = 0 \Rightarrow {C_3} = 0$

Then

$v(t) = 6\sin 2t\overrightarrow i + \left( {4\cos 2t - 4} \right)\overrightarrow j + 8{t^2}\overrightarrow k$

Let's find displacement:

$r(t) = \int {v(t)dt} = \int {6\sin 2t} dt\overrightarrow i + \int {\left( {4\cos 2t - 4} \right)dt} \overrightarrow j + \int {8{t^2}dt} \overrightarrow k = ( - 3\cos 2t + {C_4})\overrightarrow i + \left( {2\sin 2t - 4t + {C_5}} \right)\overrightarrow j + \left( {\frac{8}{3}{t^3} + {C_6}} \right)\overrightarrow k$

By condition, r(0)=0. Then

$- 3\cos 0 + {C_4} = 0 \Rightarrow - 3 + {C_4} = 0 \Rightarrow {C_4} = 3$

$2\sin 0 - 0 + {C_5} = 0 \Rightarrow {C_5} = 0$

$0 + {C_6} \Rightarrow {C_6} = 0$

Then

$r(t) = ( - 3\cos 2t + 3)\overrightarrow i + \left( {2\sin 2t - 4t} \right)\overrightarrow j + \frac{8}{3}{t^3}\overrightarrow k$

Answer: $v(t) = 6\sin 2t\overrightarrow i + \left( {4\cos 2t - 4} \right)\overrightarrow j + 8{t^2}\overrightarrow k$ , $r(t) = ( - 3\cos 2t + 3)\overrightarrow i + \left( {2\sin 2t - 4t} \right)\overrightarrow j + \frac{8}{3}{t^3}\overrightarrow k$