# Differential equation of an object dropped from certain height

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?

$$x=vt$$ 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