Physics Kinematics Problems, "Catch-up"?

Question

race car driver makes pit stop and refuels, after refueling he leaves pit with acceleration whose magnitude is 6.0 m/second squared, and after 4.0 seconds he enters the main speedway. At the same instant, another race car that is on the speedway and traveling at a constant speed of 70.0 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

Ok so i setting it up as Vo=0 for driver 1, and a=6.0m/seconds squared. and t=4 seconds. Then for driver two i got a=0 since hes going at constant speed and V=70m/s. but which formula do i use?!

3. Suppose you are visiting a planet in a distant part of the galaxy. to determine the acceleration due to gravity on this planer, u drop a rock from height of 55m. The rock strikes the ground 1.9 seconds later. How many times greater is the acceleration due to gravity on this planet than that on earth?

55=1/2(a)(1.9)squared so 3 times greater?

Answer 1

To be a good problem solver, you need to analyze the problem, isolate the core of the problem, and train and use your analytical capability.
Before the instance of the "4.0 seconds". You actually know every thing about the 1st car. His speed at the the "4.0 seconds" is 24m/s, acceleration 6.0m/s^2. Thus you may start from this instance of "4.0 seconds". Therefore Vo = 24m/s, a = 6.0m/s^2.
Let the time the first car takes to catch the other car to be t after the initial "4.0 seconds". You know that both car must cover the same distance in t seconds:
70t = Vo*t + 0.5a*t^2 = 24t + 0.5*6.0*t^2
Solve this one-variable quadratic equation you get t.

3) Remember that you have to take TWO significant figures. Thus a/g = 30.1/9.8 = 3.1
Thus the acceleration due to gravity on this planet is 3.1 times AS BIG AS that on earth. OR: Thus the acceleration due to gravity on this planet is 2.1 times GREATER THAN that on earth.