To prove!!!2

Power:

$P = F \, v = m \, a \, v \\ = 2(18)(18 t + 3) \\ = 36(18 t + 3)$

Energy:

$E = \int_0^5 P \, dt \\ = 36 \int_0^5 (18 t + 3) \, dt \\ = 36 \, (9(5^2) + 3(5)) \\ = 8640$

Since the force is constant, the definition of work as an integral is not needed. The acceleration is found by

$\Large\frac{dv}{dt}$ , which in this case is 18m/s $^2$

Newton's second law:

$\Large F=ma = 2*18=36N$

Work:

$\Large W=Fd=36d$

displacement is found by integrating velocity:

$\Large\int 18t+3$ = $\Large3t+9t^2$

substituting t=5;

$\Large d=240$

Therefore, the work done is $\Large36d=240*36=8640J$

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Displacement:

$s=\int_0^5 v\, dt\\=\int_0^5 (18t+3)\, dt\\=240$

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Acceleration:

a= $\frac{dv}{dt}$

= $\frac{d}{dt}$ (18t+3)

=18m/s $^2$

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Energy

E=Fs

=mas

=2×18×240

=8640

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