differential equations

**djmccabie** · May 13th 2009, 03:03 PM

1. A particle, of mass 0.4kg, is projected vertically upwards with an initial speed of 30 m/s. at time ts after projection, the speed of the particle is v m/s
and the magnitude of the air resistance acting on the particle is 0.2vN.

a) Show that, while the particle is ascending, v satisfies the differential equation

$\frac{dv}{dt}=-9.8-0.5v$

b) determine v in terms of t

Right so i'm a little new to differential equations so i'm unsure exactly what to do. I know that dv/dt is acceleration. I know that acceleration when the particle goes upwards verticle the force goes against gravity so thats where the -9.8 comes from. just not sure about the -0.5v

for part b i will take an educated guess that you have to integrate that equation, however its worth 6 marks so i think maybe theres more to it?

**skeeter** · May 13th 2009, 03:53 PM

Originally Posted by djmccabie

1. A particle, of mass 0.4kg, is projected vertically upwards with an initial speed of 30 m/s. at time ts after projection, the speed of the particle is v m/s
and the magnitude of the air resistance acting on the particle is 0.2vN.

a) Show that, while the particle is ascending, v satisfies the differential equation

$\frac{dv}{dt}=-9.8-0.5v$

b) determine v in terms of t

(a) while traveling upward, net force on the particle ...

$ma = -(mg + 0.2v)$

divide every term by $m$ ...

$a = -\left(g + \frac{0.2v}{m}\right)$

$m = 0.4$ kg ... $g = 9.8$ m/s^2

$\frac{dv}{dt} = -(9.8 + 0.5v)$

yes, integration is required for part (b) ...

$\frac{dv}{dt} = -(9.8 + 0.5v)$

$\frac{dv}{dt} = -0.5(19.6 + v)$

separate variables ...

$\frac{dv}{19.6 + v} = -0.5 \, dt$

integrate both sides of the equation, use your initial condition to find the constant of integration, then finalize velocity as a function of time.

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