Question

The falling object in Example 2 satisfies the initial value problem dv/dt =9.8−(v/5), v(0) =0. (a) Find the time that must elapse for the object to reach 98% of its limiting velocity. (b) How far does the object fall in the time found in part (a)?

Answer 1

Answer:

a. $t=19.56 s$

b.$d=718.34$

Explanation:

The solution to the differential equation

$\dfrac{dv}{dt}=9.8-\dfrac{v}{5}$

is the exponential function

$v(t)=ce^{-0.2t}+49$

and we find $c$ from the initial condition $v(0)=0:$

$0=ce^{-0.2*0}+49\\\\0=c+49\\\\c=-49$

Therefore, we have

$v(t)=-49e^{-0.2t}+49$

$\boxed{ v(t)=49(1-e^{-0.2t})}$

Part A:

The maximum velocity that the object can reach is 49 (which the maximum value $v(t)$ can have).

Now, 98% of 49 is 48.02; therefore,

$48.02=49(1-e^{-0.2t})$

$0.98=1-e^{-0.2t}$

$e^{-0.2t}=0.02$

$\boxed{t=19.56 s}$

Part B:

The distance traveled is the integral of the speed:

$d=\int_0^{19.56}v(t)*dt$

$d=\int^{19.56}_0 {49(1-e^{-0.2t})} \, dt$

$d=49[t+5e^{-0.2t}]_0^{19.56}$

$\boxed{d=718.34}$