Rate of change of velocity with respect to time is known as acceleration.
Position of particle is given by, [tex]s(t)=\frac{t^{3} }{3}+\frac{9}{2}t^{2}-7t+8[/tex]
Given data are, a(t) = 2t + 9, s(0) = 8, v(0) = −7
Since, Rate of change of velocity with respect to time is known as acceleration.
[tex]a(t)=\frac{dv}{dt} \\\\a(t)dt=dv\\\\(2t+9)dt=dv[/tex]
Integrating both side.
We get, [tex]v=t^{2}+9t+c[/tex]
Substituting v(0)= -7 in above equation
We get, c = -7
So, [tex]v=t^{2}+9t-7[/tex]
Since, [tex]v= \frac{ds}{dt}\\\\ds=(t^{2}+9t-7 )dt[/tex]
Integrating both side
We get, [tex]s=\frac{t^{3} }{3}+\frac{9}{2}t^{2}-7t+c[/tex]
Substituting s(0)=8 in above equation.
We get, c = 8
So, position of particle becomes,
[tex]s(t)=\frac{t^{3} }{3}+\frac{9}{2}t^{2}-7t+8[/tex]
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