Ap Physics 1 help with stacked graphs , Constant Acceleration model

Question

1. While cruising along a dark stretch of highway with the cruise control set at 25 m/s (≈55 mph), you see, at the fringes of your headlights, that a bridge has been washed out. It takes 0.50 s from the time you see the danger to apply the brakes and another 3.50 s after applying the brakes to stop. 
2. Draw an appropriate diagram of the situation. Include reference frame, values, etc. Identify the model or models (constant velocity, constant acceleration) you will apply to the situation. Be specific as to when and why you’ll apply those models. Let time zero be when you see the bridge.
3. Construct a “stack” of position, velocity and acceleration vs. time graphs that illustrate the situation described above. Graphically represent the car’s displacement, on a velocity vs. time graph, from the first hint of danger until the car was completely stopped.
4. Using the graphical representation, determine how far the car went from the time the danger was first observed until stopping was complete. (Explain your problem solving method.)
5. From the v vs. t graph, determine the acceleration of the car once the brakes were applied. (Show work.)
6. Using mathematical representation(s) of the car’s motion (appropriate for the model(s) you chose above) determine how far it went from the first hint of danger until it was completely stopped. (Show work.)

1. Now consider a slightly different situation: What if the speed of the car was 35 m/s (speeding!), and the reactions time was doubled (on a cell phone perhaps?) but the braking acceleration of the car was the same as you found in #7d above.
2. How far did the car travel from the moment the danger was seen until coming to a complete stop?
3. Solve question 8a using a different method. (Ex: if you used a graphical method, use equations.)

For each problem, provide a diagram indicating your reference frame (zero position & positive direction). 

NO DIAGRAM? NO CREDIT. If solving with graphs, include sketches of relevant graphs. If solving with equations, write the general equation before solving. It is recommended that you solve with variables and then plug in numbers at the end.

1. At t = 0 s a car has a speed of 30 m/s. After 6 s, its speed is 15 m/s. If acceleration is constant, how far does the car travel during the 6 seconds? If the motion doesn’t change, how far does the car travel until it stops?
2. A car whose initial speed is 30 m/s speeds up uniformly to 50 m/s in 5 seconds. 
3. What is the acceleration of the car? 
4. How far does it travel in the 3rd second (from t = 2 s to t = 3 s)?
5. A physics student skis a distance of 150 meters down a hill, accelerating at a constant 2.5 m/s2. What is her velocity at the bottom if she started from rest at the top of the hill?
6. A jet needs a speed of 75 m/s to take off. If the runway is 1600 m long, what acceleration does it need to attain takeoff speed while still on the runway?
7. A subway train in Washington D.C. starts from rest and accelerates at 2.0 m/s2 for 12 s.
8. How far does the train travel?
9. What is the train’s velocity after 12 s?
10. A car traveling at 15 m/s slams on the brakes to avoid an accident. The car skids 25 m before stopping. What is the acceleration of the car?

Answer 1

8th last question/problem

15 m/sec = initial speed = vo, the car skids 25 meters before stopping. stopping means velocity = 0

general formula is

s(t) = (a/2)t^2 + vot +so where so=initial distance = 0, final distance = 25 meters

t= time measured in seconds. a=acceleration <0, assuming a is a constant

s(t) =25= (a)t^2/2 +15t

a = v'(t) = s''(t)

v=s'(t) = at+15

v=0 when a=15/t

s(t)=25 =(15/t)t^2/2+15t

50=15t+30t = 45t

t = 50/45 = 10/9 second to stop

a= 15/(10/9) = 15(9/10) = (3/2)9 = 13 1/2 meters per second per second = acceleration to go from s=0 to s=25

in 1 second, it slows down from 15 to 1.5 m/s, in 10/9 seconds= 1.111... seconds, it skids to a stop when v = 0

so a= -13 1/2 m/sec^2 to go from 25 to 0, from vo=15 to v=0, a has to be negative to slow down & stop