Calculate initial velocity based on displacement, time and constant acceleration.

"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.

• So, what equations are you allowed to use? The two facts translate into a fairly straightforward system of two equations in two unknowns. – amd Apr 19 at 18:44
• @amd what would the system of equations be? I was trying something with integrals. – muhammad haider Apr 19 at 18:55
• The first one is something like $v+50a=2v$, where $v$ is the initial velocity and $a$ the acceleration. The second would use the total travel time and distance to relate $a$ and $v$. If need be, you can derive that one by integrating velocity w.r. time. – amd Apr 19 at 19:10

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$$^2$$.