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Q11P

Expert-verifiedFound in: Page 117

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A ${\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{kg}}$**** particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by role="math" localid="1657014461331" ${\mathit{x}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{0}}{\mathbf{m}}{\mathbf{+}}{\mathbf{(}}{\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}{\mathbf{)}}{\mathit{t}}{\mathbf{+}}{\mathit{c}}{{\mathit{t}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{(}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathit{s}}{\mathbf{)}}{{\mathit{t}}}^{{\mathbf{3}}}$**

The value of constant c is $+9.0$

- Mass of particle, $m=2.0\mathrm{kg}$
- Position of a particle in expression, $x=3.0+4t+c{t}^{2}-2{t}^{3}$
- Magnitude of force at
*t*= 3.0 s, $F=-36\mathrm{N}$ which is in a negative direction

**The acceleration of the object is equal to the time rate of change of velocity. The acceleration can be found by differentiating the velocity vector with respect to time or differentiating the displacement vector twice with respect to time.**

**As we are given the function for the position of a particle if we differentiate that for time, we will get twice the acceleration. As we have got the value of force at a particular time, using Newton’s 2nd law, we can find the value for c.**

Formula:

The force on a body due to Newton’s second law, $\overrightarrow{F}=m\overrightarrow{a}$ (i)

Here, $\overrightarrow{F}$ is net force, is mass of the object, and $\overrightarrow{a}$ is the acceleration.

The acceleration of a body in motion, $a=\frac{{d}^{2}x}{d{t}^{2}}$ (ii)

Taking the derivative of the given position equation$x=3.0+4t+c{t}^{2}-2{t}^{3}$ with respect to time, we get the velocity as follows:

role="math" localid="1657015128101" $v=0+4+2ct-6{t}^{2}..............\left(a\right)$

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