Answer :
To solve this problem we will apply the linear motion kinematic equations. For which the change of the initial and final velocity is described, as the product of twice the acceleration and the distance traveled. This equation is independent of time, so it is conducive to this exercise.
We will start converting the units.
[tex]v_0 = 25km/h (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]v_0 = 6.544m/s[/tex]
The definition of the linear equation of kinematic motion states that
[tex]v_f^2 - v_0^2=2ax[/tex]
Here,
[tex]v_f[/tex] = Final velocity
[tex]v_0[/tex] = Initial velocity
a = Acceleration
x = Displacement
Rearranging to find the acceleration,
[tex]a = \frac{v_0^2}{2x}[/tex]
Replacing,
[tex]a = \frac{6.544}{2(15)}[/tex]
[tex]a = 1.61m/s^2[/tex]
Now we know that speed is
[tex]v_0 = 50 \frac{km}{h} (\frac{1000m}{1km})(\frac{1h}{3600s})[/tex]
[tex]v_0 = 13.889m/s[/tex]
Then the distance traveled, starting from the same previous equations would be
[tex]x = \frac{v_0^2}{2a}[/tex]
[tex]x = \frac{(13.889)^2}{2(1.61)}[/tex]
[tex]x = 60m[/tex]