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Sagot :
(a). Given data:
* The velocity of the particle in the negative x-direction is,
[tex]v_i=-4.36ms^{-1}[/tex]* The velocity of the particle in the positive x-direction is,
[tex]v_f=7.94ms^{-1}[/tex]* The time taken by the particle is,
[tex]t=10.9\text{ s}[/tex]Solution:
The acceleration in terms of velocity and time is,
[tex]a=\frac{v_f-v_i}{t}[/tex]Substituting the known values,
[tex]\begin{gathered} a=\frac{7.94-(-4.36)}{10.9} \\ a=\frac{7.94+4.36}{10.9} \\ a=1.128ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the particle is 1.128 meter per second squared.
(b). The velocity before the 10.9 second of the -4.36 m/s velocity is,
[tex]\begin{gathered} a=\frac{v_f-v}{t} \\ 1.128=\frac{-4.36-v_{}}{10.9} \\ 12.295=-4.36-v \\ -v=12.295+4.36 \\ -v=16.655 \\ v=-16.655ms^{-1} \end{gathered}[/tex]Thus, the value of the velocity is -16.655 meter per second.
(c). Let at time t particle will come to rest,
The time t in terms of acceleration is,
[tex]\begin{gathered} a=\frac{v_f-v_i}{t} \\ 1.128=\frac{0-(-4.36)_{}}{t} \\ t=\frac{4.36}{1.128} \\ t=3.865\text{ s} \end{gathered}[/tex]Thus, the particle will come to rest after 3.865 s since the velocity 4.36 m/s in negative x-axis.
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