# Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 133: 29

$v=2.01ft/s$

#### Work Step by Step

The required velocity can be determined as $\Sigma F_y=0$ $\implies F-W=ma$ We plug in the known values to obtain: $100t-200=\frac{200}{32.2}a$ $\implies a=16.1t-32.2$ Now, $v=\int_2 ^t (16.1t-32.2)dt=(\frac{16.1}{2}t^2-32.2t)|_2^t$ $\implies v=8.05t^2-32.2t+32.2$ This simplifies to: $v=2.01ft/s$

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