## Related questions with answers

The function $\vec{p}=-3\left(2^t\right) \hat{i}+4 \cos (2 t) \hat{j}$ describes the position of a particle in the xy-plane. What is the speed and the magnitude of the acceleration vector of the particle when t=2?

Solution

VerifiedWe have the position vector is

$\overrightarrow{p(t)}=-3(2^t)i+4 \cos(2t)j$

To find the velocity , we differentiate the componants of $\overrightarrow{p(t)}$

Thus, the velocity vector is

$\color{#4257b2}\overrightarrow{v(t)}=-3\dfrac{d}{dt}(2^t)i+4\dfrac{d}{dt}[ \cos(2t)]j=-3\ln 2\cdot 2^ti-8\sin(2t)j$

To find the acceleration , we differentiate the componants of $\overrightarrow{v(t)}$

Thus, the acceleration vector is

$\color{#4257b2}\overrightarrow{a(t)}=\dfrac{d}{dt}(-3\ln 2\cdot 2^t)i-8\dfrac{d}{dt}[ \sin(2t)]j=-3(\ln 2)^2 2^ti-16\cos(2t)j$

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