Slope
 
Reply Mon 25 Feb, 2013 11:48 am
Hello all, so I am retaking an intro to physics course in preparation for my mcat. This is a new school I am taking the course at and while I seem to understand physics(in my opinion), ive just never seen physics questions phrased this way someone please help because I am lost.

1)An object moves from the position r1=(1,3,-5) to the position r2(-1,4,8) during the time 8 s. Find: (a) displacement; (b) distance; (c) average velocity; (d) average speed.

2)A car is traveling a road that makes quarter of a circle with a constant speed s. What is the magnitude of the average velocity v of the car?

3. Two stones are thrown vertically up with the initial velocity v0, second stone with a time delay t0 with respect to the first one. At which time after the beginning of the motion the two stones will collide? Is this scenario possible for any v0 and t0? What are the limitations?

4. A gunman shoots from a riffle in a horizontal direction without correcting for gravity. How much below the intended target, at a horizontal distance d, will the bullet strike if its initial speed is v0?

5. A ship crosses a river aiming at the angle (theta) to the left from the straight course. The speed of the ship with respect to water is v ́. The width of the river is d and the water velocity is u to the right. What will be the side displacement h of the ship as it lands on the other shore?
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Question • Score: 0 • Views: 1,401 • Replies: 1
No top replies

 
fresco
 
  1  
Reply Mon 25 Feb, 2013 01:07 pm
@Slope,
SOME HINTS

1. Distance traveled between (a,b,c) and (p,q,r) is given by
d = sqrt [(a-p)^2 +(b-q)^2 + (c-r)^2] PYTHAGORAS

3. Equate the heights to get t in terms of to ( by using h=ut-0.5 gt^2)

4. Horizontally t=d/v0. Vertically h=0.5 gt^2

0 Replies
 
 

Related Topics

Rotational Kinematics HW Help - Question by jdmaxwell02
Is my solution right? - Question by greatestever298
 
  1. Forums
  2. » Physics
Copyright © 2024 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 04/20/2024 at 01:37:21