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Q. 40

Expert-verifiedFound in: Page 61

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A particle’s velocity is described by the function vx = kt 2 m/s, where k is a constant and t is in s. The particle’s position at t0 = 0 s is x0 = - 9.0 m. At t1 = 3.0 s, the particle is at x1 = 9.0 m. Determine the value of the constant k. Be sure to include the proper units

The value of constant $k$ is 2.

The velocity of a particle is given by a function ${v}_{x}=k{t}^{2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The position of the particle at ${t}_{0}=0s$ is ${x}_{0}=-9.0m$.

The position of the particle at ${t}_{1}=3.0s$ is ${x}_{1}=9.0m$.

The position of a particle is given by

${x}_{f}={x}_{0}+{\int}_{{t}_{0}}^{{t}_{f}}{v}_{x}dt$

here,

role="math" localid="1648490216091" ${x}_{1}={x}_{0}+{\int}_{{t}_{0}}^{{t}_{1}}{v}_{x}dt\phantom{\rule{0ex}{0ex}}={x}_{0}+{\int}_{{t}_{0}}^{{t}_{1}}k{t}^{2}dt\phantom{\rule{0ex}{0ex}}={x}_{0}+{\left[\frac{1}{3}k{t}^{3}\right]}_{{t}_{0}}^{{t}_{1}}\phantom{\rule{0ex}{0ex}}={x}_{0}+\frac{1}{3}k{t}_{1}^{3}-\frac{1}{3}k{t}_{0}^{3}$

Substitute known values

$9=-9+\frac{1}{3}k{\left(3\right)}^{3}-\frac{1}{3}k{\left(0\right)}^{3}\phantom{\rule{0ex}{0ex}}9=-9+9k\phantom{\rule{0ex}{0ex}}9k=18\phantom{\rule{0ex}{0ex}}k=2$

Therefore, the value of constant $k$ is 2.

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