A force f = (4−v²) N is acting on a particle of mass 3kg where v is velocity in ms⁻¹ of the particle. Initially, the particle is at rest at the origin. Find, i. The time taken for the particle to reach a velocity of 0.5 ms⁻¹ ii. The displacement when the velocity is 0.5 ms⁻¹

$$F = m \frac{dv}{dt}$$ $$4-v^2 = 3 \frac{dv}{dt}$$ $$\frac{dv}{dt} = \frac{4-v^2}{3}$$

$$dt = \frac{3}{4-v^2} dv$$

Integrate from $t=0$ to $t$ and $v=0$ to $v=0.5$:

$$\int_{0}^{t} dt = \int_{0}^{0.5} \frac{3}{4-v^2} dv$$ $$t = 3 \int_{0}^{0.5} \frac{1}{2^2-v^2} dv$$

Using the standard integral $\int \frac{1}{a^2-x^2} dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right|$, with $a=2$:

$$t = 3 \left[ \frac{1}{2(2)} \ln \left| \frac{2+v}{2-v} \right| \right]_{0}^{0.5}$$ $$t = \frac{3}{4} \left[ \ln \left( \frac{2+v}{2-v} \right) \right]_{0}^{0.5}$$

$$t = \frac{3}{4} \left( \ln \left( \frac{2+0.5}{2-0.5} \right) - \ln \left( \frac{2+0}{2-0} \right) \right)$$ $$t = \frac{3}{4} \left( \ln \left( \frac{2.5}{1.5} \right) - \ln(1) \right)$$ $$t = \frac{3}{4} \ln \left( \frac{5}{3} \right)$$ $$t \approx \frac{3}{4} \times 0.5108256$$ $$t = \boxed{0.383 \, \text{s}}$$

$$F = m v \frac{dv}{dx}$$ $$4-v^2 = 3 v \frac{dv}{dx}$$

$$dx = \frac{3v}{4-v^2} dv$$

Integrate from $x=0$ to $x$ and $v=0$ to $v=0.5$:

$$\int_{0}^{x} dx = \int_{0}^{0.5} \frac{3v}{4-v^2} dv$$ $$x = \int_{0}^{0.5} \frac{3v}{4-v^2} dv$$

Use substitution: Let $u = 4-v^2$. Then $du = -2v \, dv$, so $v \, dv = -\frac{1}{2} du$.
When $v=0$, $u=4$. When $v=0.5$, $u=4-(0.5)^2 = 3.75$.

$$x = \int_{4}^{3.75} 3 \left( -\frac{1}{2} \frac{1}{u} \right) du$$ $$x = -\frac{3}{2} \int_{4}^{3.75} \frac{1}{u} du$$

$$x = -\frac{3}{2} [\ln|u|]_{4}^{3.75}$$ $$x = -\frac{3}{2} (\ln(3.75) - \ln(4))$$ $$x = -\frac{3}{2} \ln \left( \frac{3.75}{4} \right)$$ $$x = -\frac{3}{2} \ln \left( \frac{15}{16} \right)$$

Using the logarithm property $\ln(a/b) = -\ln(b/a)$:

$$x = \frac{3}{2} \ln \left( \frac{16}{15} \right)$$ $$x \approx \frac{3}{2} \times 0.0645385$$ $$x = \boxed{0.0968 \, \text{m}}$$

Dernière correction gratuite aujourd'hui — reviens demain, ou tape /upgrade pour l'illimité.