Can someone double check my answers to this practice exam?

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Homework Statement



http://www.physics.rutgers.edu/ugrad/123/prevExams/CH1_2014.pdf

Homework Equations



basically just kinematics, projectile motion, and force equations.[/B]

The Attempt at a Solution



I got :
1. either B or D
2. D
3.B
4.A
5.E
6.C
7.D
8.A
9.B
10.B
11.A
12.C
13.B
14.E
15.C
16.A
17.A
I was using this practice exam to study but quickly realized there were no answers posted, so I'm unaware if what I'm doing is correct..
Thank you! [/B]
 
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haruspex said:
I get different answers for 1, 10, 11, 13, 14, 16.
okay, so for 1. I think it's B because A can't be true since part of the slope is negative and thus negative velocity. C can't be true because at t=1 the velocity is positive, and t=7 the velocity is negative. D can't be true because the acceleration is zero at that point because the position plot is a line without a changing slope. E can't be true because at t=6, the ball has a negative velocity.

Now ,
10. The helicopter goes up 1000 m. Now it begins to move 45 degrees at an acceleration of 2 m/s^2 for 30 seconds.. so final velocity= 2(30). to find avg velocity, i did (vfinal-vinitial)/2 which brings us to 30. so 30 m/s * 30 seconds = 900 ft..so i did 900+1000 = 1900..i think i went wrong somewhere calculating velocity because Vinitial might not have been 0 as the copter was already in motion but idk how to find vinitial..

11. i don't even know how to approach this question. i suppose its something to do with projectile motion definitions.. maybe to do with the the independence of x and y components

13. Now i got D. i got this because i re-read the question, it says "relative to the water", so I just divided .20 km by 8 m/s to see how long it would take.

14. I got D, i just added the components of the two forces and solved for magnitude.

16. now i got B, realized i plugged into equation incorrectly..
 
RUphysics3 said:
okay, so for 1. I think it's B because A can't be true since part of the slope is negative and thus negative velocity. C can't be true because at t=1 the velocity is positive, and t=7 the velocity is negative. D can't be true because the acceleration is zero at that point because the position plot is a line without a changing slope. E can't be true because at t=6, the ball has a negative velocity.

Now ,
10. The helicopter goes up 1000 m. Now it begins to move 45 degrees at an acceleration of 2 m/s^2 for 30 seconds.. so final velocity= 2(30). to find avg velocity, i did (vfinal-vinitial)/2 which brings us to 30. so 30 m/s * 30 seconds = 900 ft..so i did 900+1000 = 1900..i think i went wrong somewhere calculating velocity because Vinitial might not have been 0 as the copter was already in motion but idk how to find vinitial..

11. i don't even know how to approach this question. i suppose its something to do with projectile motion definitions.. maybe to do with the the independence of x and y components

13. Now i got D. i got this because i re-read the question, it says "relative to the water", so I just divided .20 km by 8 m/s to see how long it would take.

14. I got D, i just added the components of the two forces and solved for magnitude.

16. now i got B, realized i plugged into equation incorrectly..

okay so..
10. found it somewhere online. i was calculating it where it was 1000 feet straight up and then 900 ft in a direction that was 45 degrees clockwise to that, but apparently it's supposed to be a 90 degree turn then 900 feet. so then its just sqrt(1000^2+900^2)=1350 (A)
11. Okay, so h=1/2gt^2. so as max height increases, so does time in air. this means that C must be correct!
 
RUphysics3 said:
okay so..
10. found it somewhere online. i was calculating it where it was 1000 feet straight up and then 900 ft in a direction that was 45 degrees clockwise to that, but apparently it's supposed to be a 90 degree turn then 900 feet. so then its just sqrt(1000^2+900^2)=1350 (A)
11. Okay, so h=1/2gt^2. so as max height increases, so does time in air. this means that C must be correct!
All good now except for 13.
 
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