The acceleration of a particle is given by $a_x(t)=-1\mathrm{.}98\frac{m}{s^2}+(2\mathrm{.}95\frac{m}{s^3})t.$

Part A

Find the initial velocity $v_{0x}$ such that the particle will have the same $x-$coordinate at time $t=3\mathrm{.}96s$ as it had at $t=0.$ Express your answer in meters per second.

$v_{0x}=$

$\frac{m}{s}$

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Part B

What will be the velocity at time $t=3\mathrm{.}96s?$

Express your answer in meters per second.

$v_x={\displaystyle \underset{\phantom{}}{\overset{m}{}}}\frac{m}{s}$

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