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I want to solve this problem -

A ball of mass 2kg is dropped from a tall building with zero initial velocity. In addition to gravity, the ball experience a damping force of the form -2v, where v is the instantaneous velocity. Given $$ g=10m/s^2 $$ find the distance travelled by the ball in t second.

To solve the problem I have set up the differential equation as $$ m \frac {dv}{dt}= mg-2v $$

Now by solving this equation I have calculated velocity as

$$ v = 10 + ce^{-t} $$ From initial condition that the ball is dropped with zero initial velocity, the integration constant c comes out to be -10 $$ v = 10 - 10e^{-t} $$

Now to calculate the distance travelled by the ball in t time, I have to solve this differential equation for x

$$ dx/dt = 10-10e^{-t} $$

My question is why can't I use$$ distance=speed ×time$$ and calculate distance travelled directly as $$x = vt = (10 -10e^{-t})t$$ I know this answer is wrong but I want to know why?

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$x=vt$ works if $v$ is constant in the time interval $t$.
The expression $dx=v$ $dt$ works because you can regard v as constant in the infinitesimal time span $dt$, resulting in an infinitesimal displacement $dx$.
Summing up these displacements means taking the integral $$\int dx=\int v(t)dt$$ where $v(t)$ is the velocity you found as a function of time. Because $v$ is not constant in the integration range, you can't simply multiply it by it

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  • $\begingroup$ Thanks. I understand now. $\endgroup$ May 25, 2023 at 10:40
  • $\begingroup$ Dinesh now prove why if a ball falls onto an elastic material, the ratio of drop height, h to stop depth, d h/d = a, the shock level of acceleration [g's].. I used this fact and proved with accelerometers for drop tests on products. $\endgroup$ May 25, 2023 at 12:36

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