(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

1. TWO ELEVATORS OPERATING IN PARALLEL SHAFTS, APPROACH EACH OTHER FROM POSITIONS WHICH ARE 120m APART. THE UPPER CAR HAS A DOWNWARD ACCELERATION OF 0.5m/s^2 AND THE LOWER CAR IS BEING ACCELERATED UPWARD AT 1m/s^2. WHEN AND WHERE WILL THEY PASS EACH OTHER IF THE LOWER CARD STARTS 1 SECOND AFTER THE OTHER CAR.

2.A BODY MOVES IN A STRAIGHT LINE SO THAT ITS ACCELERATION IS IS a=3v, WHERE "a" IS IN m/s^2 AND "v" in m/s. IF s=3m AND v=1.2m/s, FIND

A) "v" IN TERMS OF "s"

B)THE VALUES OF VELOCITY AND ACCELERATION WHEN s = 10m.

3)A FLYWHEEL 0.8m IN DIAMETER ACCELERATES UNIFORMLY FROM REST TO 1200 RPM IN 30 SECONDS. HOW MANY REVOLUTIONS DOES THE FLYWHEEL MAKES IN ATTAINING ITS SPEED OF 1200 RPM?

4) AN AUTOMOBILE WEIGHING 9KN TRAVELS AT 80km/hr. AT THE FOOT OF AN INCLINE OF 10 DEGREES, THE MOTOR IS TURNED OFF. HOW FAR DOES THE CAR TRAVEL UP THE INCLINE BEFORE STOPPING?

5) CARS A AND B APPROACH EACH OTHER FROM POINTS 460m APART. CAR A HAS AN INITIAL SPEED OF 70km/hr AND IS DECELERATING AT THE RATE OF 0.40m/s^2. B HAS AN INITIAL SPEED OF 20km/hr. AND IS ACCELERATING AT THE RATE OF 0.30m/s^2. WHEN WILL THE CARS MEET AND HOW FAR WILL EACH HAVE TRAVELLED?

2. Relevant equations

3. The attempt at a solution

FOR NUMBER 1) I ASSUMED THAT THE TOTAL DISTANCE OF 120 IS EQUAL TO D(elevator A)+D(elevator B)....now for the time...TIME ELEVATOR A = t and TIME ELEVATOR B = t+1....after that I'm lost....

FOR NUMBER 2) NO IDEA

FOR NUMBER 3) NO IDEA

FOR NUMBER 4) NO IDEA

FOR NUMBER 5) I SET THE THE TOTAL DISTANCE 460m = D(car A) + D(car B), THEIR TIME IS EQUAL.... I USED THE KINEMATIC EQUATION "s = Vot + 1/2at^2"

460 = Vot(car A) + Vot(car B) - 1/2(a)(t^2)(Car A) + 1/2(a)(t^2)(Car B)

I used the quadratic formula and arrived at

t = 19.13 seconds

D(car A) = 298.69 m

D(car B) = 161.31 m

ANY HELP AND SUGGESTIONS WILL BE GREATLY APPRECIATED.

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# Homework Help: Problems concerning dynamics

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