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JustRandomThings · Mar 13, 2017

Hi everyone thanks in advance!

A particle moves along a straight line and its acceleration when it is moving at velocity v m/s is numerically equal to a=-(100+v^2) m/s^2. Its initial speed is 10 m/s

1. Find the distance travelled before the particle comes to a rest

2. Find the time taken before the particle comes to rest

Mahan1 · Mar 26, 2017

JustRandomThings said:
Hi everyone thanks in advance!

A particle moves along a straight line and its acceleration when it is moving at velocity v m/s is numerically equal to a=-(100+v^2) m/s^2. Its initial speed is 10 m/s

1. Find the distance travelled before the particle comes to a rest

2. Find the time taken before the particle comes to rest

$\textrm{let} \ t= t_{1}, x =x_{1} \ \textrm{such that } \ v_{t_{1}} = v_{x_{1}} = 0$

note
$a= \frac{dv}{dt} = v \frac{dv}{dx}$

to get the distance:
$a= v \frac{dv}{dx} = -(100 + v^{2}) \Rightarrow \frac{dv}{dx} = - \frac{100}{v}- v$

$\Rightarrow \int_{0}^{x_{1}} \frac{v'}{\frac{100}{v} + v } dx = - \int_{0}^{x_{1}} 1 dx \ \textrm{where} \ \ v' := \frac{dv}{dx}$

$\Rightarrow \int_{0}^{x_{1}} \frac{vv'}{100+v^{2}} dx = \frac{1}{2}( ln(100+v_{x_{1}}^{2}) - ln(100 + v_{0}^{2})= - x_{1}$

$\Rightarrow \frac{1}{2}ln(\frac{1}{2}) = -\frac{1}{2}ln(2) = -x_{1} \Rightarrow \frac{1}{2}ln(2) = x_{1}$

since a < 0 and we are calculating the distance when the object comes into the rest then there is no back tracking therefore the total distance travelled is $x_{1}$

Using the other equality for the acceleration:

$a = \frac{dv}{dt} = -100 -v^{2}$

$\Rightarrow \int_{0}^{t_{1}} \frac{v'}{100 +v^{2}} dt = -\int_{0}^{t_{1}} 1 dt$ where $v' = \frac{dv}{dt}$

$\Rightarrow tan^{-1} (\frac{v_{t_{1}}}{10} )-tan^{1} (\frac{v_{0}}{10}) = - t_{1}$

$\Rightarrow -tan^{-1}0 - tan^{-1} (1) = -t_{1}$
$\therefore t_{1} = \frac{\pi}{4}$

Drongoski · Mar 27, 2017

$t_1 = \frac {\pi}{40}$

??

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JustRandomThings

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