"A car has a constant speed along a road. It goes down a hill at a constant acceleration. 50s after it goes down the hill the speed is doubled and 50s later it reaches the end of the 200m hill and is back at a constant speed. Find out the initial velocity and acceleration."
At first I made relevant graphs to see if I could find some useful information from that but no luck. Then I tried to use the "suvat" equations but we haven't learned them in class so I'm not allowed to use them, which is why I'm stuck as to how to solve this basic problem.
Using what @amd said,
The first equation is $v+50a=2v$, because it has constant acceleration of $a$ m/s$^2$.
Therefore, by the end of the hill, it has speed $2v+50a=3v$.
Now, knowing that in $100$ seconds, starting at an initial velocity of $v$, and ending at a velocity of $3v$, the car traveled $200m$.
Therefore, the average speed of the car was $2v$, which is $2$ m/s.
So that means that the initial velocity is $1$ m/s.
In addition, since the car took $50$ seconds to accelerate from $1$ m/s to $2$ m/s, the acceleration is $0.02$ m/s$^2$.