Please help for Physics 201!?

Question

How do you solve these problems? I tried several times, but I just can't get the right answers! (cries)

1. Two cyclists start from rest 73.1m apart and ride toward each other. Cyclist One's acceleration has magnitude of 0.236m/s^2. Cyclist Two's acceleration has magnitude of 0.252m/s^2. Relative to Cyclist One's starting point, where do they collide?

2. A pellet gun is fired straight downward from the edge of a cliff that is 18.1m above ground. The pellet strikes ground with speed 39.8m/s. How far above cliff edge would pellet have gone had gun been fired straight upward?

3. A race driver has made pit stop. After, he leaves pit with an acceleration of 7.25m/s^2, and after 5.05s he enters the main speedway. At the same instant, another racer on the speedway traveling 78.0m/s overtakes and passes entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

Thank you to anyone who can help!!

Answer 1

1.

Ok, first, let the distance they collide from cyclist one to be x. Logically, the distance which they collide from cyclist two would be 73.1-x.

Next, use the simple kinematics formula, which is S = ut + 0.5 a t^2. Where S is the distance travelled, u is the initial speed, and a represents the acceleration. For both u's, it is 0 as they start from rest. Now, you'd probably get two equations with x in them. No fear, because you're on the right track. Simply solve them simultaneously, to get the value of x which should be 35.3m.

2.

For this question, another valuable equation is used. v^2 = u^2 + 2as, with all unknowns being as above and v being the final speed.

Again, the data you have is that the cliff is 18.1m high, and the final velocity is 39.8. If we simply take downward vectors to be negative, we'd get 1584.04 = u^2 + 2(-9.81)(-18.1)

Solving for u, you'd get u = 35.06 m/s.

Now to find the highest point of the pellet. What we do know from a projectile motion is that at the highest point, the vertical component of velocity is 0. Now simply use the same equation as above, and you'd get 0 = 35.06^2 + 2(-9.81)(s).

Solve for s, and you'd get s = 62.64m.

3.

After 5.05s, the entering car has a velocity of 36.6 m/s. Now simply set the distance which both cars would travel to be x. Assuming that the overtaking car travels at constant velocity, and using the equation s = ut + 0.5(a)(t^2), put all the values you have of the overtaking car in.

Thus, x = 78.0(t), since we have already assumed 0 acceleration.

Seeing as the entering car also travels x m to catch up and overtake, we have:

x = (36.6)(t) + 0.5(7.25)(t^2).

Now simply solve for t by placing 78.0t in the place of x, and you'd get

t = 11.42s. Thus that is the amount of time required to catch the other car.

I hope this helps you!