## End of Chapter 1 Physics

chibean
Posts: 11
Joined: Fri Jan 03, 2020 12:10 pm

### End of Chapter 1 Physics

For end of physics chapter 1 question #6 can you explain the equations used for each step and the steps needed to solve for this equation? I thought using the equations on page 15 figure 9 would help solve the problem but the answer explanation only describes using the main 4 kinematics equation.

Also which equation was used to help solve for time?
chibean
Posts: 11
Joined: Fri Jan 03, 2020 12:10 pm

### Re: End of Chapter 1 Physics

Also for question # 7 I selected A because constant velocity means there is no acceleration but the correct answer was C and the explanation discussed net force but the question did not discuss force so I was confused on why force was related to the question. Can you further explain?
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

### Re: End of Chapter 1 Physics

#6:
The correct answer is D, and I believe the book states this too. You are correct in saying A, but both A and B are true, so answer choice D is correct. The acceleration of the car is constant and zero.

#7:
There are a bunch of ways to solve this, but the most straightforward is to just use vf^2 = vi^2 + 2ad, seeing as we have all the necessary ingredients for that (we have a final velocity, we have a distance and an acceleration - leaving us with only one unknown).

So, let's rearrange:

Code: Select all

``````vf^2 - 2ad = vi^2

(20^2-4*75)^0.5 = vi

(400-300)^0.5 = vi

10 = vi
``````
And that is it. Other expressions work too, and depending on which you use you might end up with a quadratic to solve for time. But this is certainly the recommended way.
chibean
Posts: 11
Joined: Fri Jan 03, 2020 12:10 pm

### Re: End of Chapter 1 Physics

Thank you, but I think you confused my questions lol.

But I meant #6 & #7 from the passage questions.

so #7
Patients moved through the dart throwing motions described in figure 2 with a constant rotational velocity. Given this, which of the following statements is true?

and #6
A patient is asked to throw a dart in order to hit the bullseye in the center of a target 5 m away. If the initial velocity of the dart is 10 m/s at 45° with the horizontal, from a height level with the bullseye, where will the dart hit the target?
NS_Tutor_Mathias
Posts: 616
Joined: Sat Mar 30, 2019 8:39 pm

### Re: End of Chapter 1 Physics

For #6:
viewtopic.php?f=12&t=2667&sid=69958a68d ... daea46bf91

For #7:
This is an oddly designed question and has a few flaws, but what it is trying to tell you is that if theta = 30 degrees, the horizontal component of the tension vector in the scapholunate will be half the total vector (by trigonometric properties: sin 30 = 0.5).

Therefore, for the horizontal forces to balance, the total tension in the scapholunate must be twice that in the radioscaphocapitate.

The big logical flaw here is of course that in a real system, the contact forces (normal reaction forces) between bones is really what keeps them stationary - tension on the ligaments does not necessarily have to all balance out. That would be impossibly unlikely, seeing as we can remain static in a variety of positions with very different angles between all our bones and ligaments.