After some reflection, I found that t could be expressed as a function of A using the equation of v(t) :
v(t) = 0 --> I - A.t = 0 --> 25 - A.t = 0 --> t = 25/A
Substituting 25/A for t in the last equation alllows to find the desired answer given in the key : A = 25/4 = 6,25.
What astonishes me is that the constant of integration plays no role here. The primitive I used did not take into account any constant of integration; however, it works. How comes?
( Source of the exercice: http://www.buders.com/UNIVERSITE/Universite_Dersleri/Math101/Arsiv/integral_sorulari_ve_cozumleri.pdf)
I'm working on Exercice 20 ( below) . Here is what I have done so far, but I'm stuck.
- The speed function v is ( with I = initial speed, and A = deceleration rate) :
v(t) = I - A.t
- The distance function d should be the integral of v, and hence a primitive of v
d(t)= I.t - (A/2)t²
With I = 25m/s, we are given the output of 50m
So the problem results in solving :
25t - (A/2)t²= 50
which can be rearranged as :
-(A/2)t² + 25 t -50 = 0
But that must be wrong, since I guess I cannot find A ( deceleration rate), t being also unknown.
Could you please tell me at which point I got on the wrong track?
The condition $d(0)=0$ sets the integration constant $C$ in $D=C+It-At^2/2$ to $0$. Your approach worked not because $C$ doesn't matter, but because it really is $0$. To solve, first note the stopping time satisfies $0=I-At$. So $D=At^2/2\implies A=2D/t^2$. See also here.