Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Fundamental Problems - Page 16: 6

Answer

The velocity of the particle when s = 10 m is equal to 14.3 m/s

Work Step by Step

1. Using equation 12-3: $$a \space ds = v \space dv$$ 2. Integrate both sides and substitute the values and expressions: $$\int_0^sa \space ds = \int_{v_0}^vv \space dv$$ $$\int_0^s (10 - 0.2s) \space ds =\Big[ \frac{v^2}{2} \Big]_{v_0}^v$$ $$\Big[10s - 0.1s^2 \Big]^s_0 = \frac{v_2}{2} - \frac{v_0^2}{2}$$ $$\Big[10s - 0.1s^2 \Big]^{10}_0 = \frac{v_2}{2} - \frac{(5)^2}{2}$$ $$10(10) - 0.1(10)^2 = \frac{v^2}{2} - 12.5$$ $$90 = \frac{v^2}{2} - 12.5$$ $$90 + 12.5 = \frac{v^2}{2}$$ $$v = \sqrt{2(102.5)} = 14.3 \space m/s$$
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