23
qwerty
·2010-10-01 01:55:17
1> wen police reached the point culprit already travelled 20 x 10 = 200 m .
now suppose u r with the culprit. wat u see is u r at rest , police is 200 m away frm u , and police has 5 speed m/s
so time taken by police to reach the culprit after turning pt = 200/5=40 sec
thus dist travelled by police = 25 x 40 = 1000 m
2>
relative speed = 18kph = 5 m / s
now suppose u r in the car with speed 42 kph
u see that u r at rest and a car is overtaking u at a speed of 5 m /s .
now the car will hav to travel 10 m to completely overtake the other car so time taken = 10/5 = 2 sec
hence distance travelled wrt grnd=time x speed wrt grnd= 2 x 60 x 5 / 18 m
basically we r using the reference frame just to find out the time , and since time is independent of frame it can u used to find distance travelled in grnd frame
1
sanchit
·2010-10-01 02:07:39
thnxx alot bhaiiya ......
1
sanchit
·2010-10-01 02:14:46
3)the acc of a particle as seen from two frames S1 S2 have equal magnitudes 4 m/s^2
a)the frames must be at rest w.r.teach other.
b)the frame may be moving w.r.teach other but neither shud be acc. w.r.t. the other
c)the acc. of s2 w.r.t. s1 may either be 0 or 8 m/s^2
d)the acc. of s2 w.r.t. s1 may be anything b/w 0 nd 8m/s^2
ans given is d i was thinkin tht ans is c..
1
Euclid
·2010-10-01 02:17:16
think...think...nice question.....ans given is correct
1
sanchit
·2010-10-01 02:19:00
bhaiiya meene socha thtss why posted
23
qwerty
·2010-10-01 04:46:55
sanchit 1 hint for u :
it is no where mentioned in the question that s1,s2,the particle hav to move along a straight line ,
dont restrict it to only 1D
it may be 2D, or even 3D [3]
1
Euclid
·2010-10-01 18:25:17
assume that particle is stationary...but S1 and S2 moving with equal accln of 4m/s2, one along y-dirn and other along x-dirn....
abhi toh relative accln of S2 wrt S1 is neither 0 nor 8!!!
1
sanchit
·2010-10-01 22:33:40
3)the distance b/w two moving particles at any time is a.if v be the relative vel. nd v1 nd v2 be the components of v along nd perpendicular to a .find the time when they are closest to each other??? nd the minimum distance between them ?????
1
nihal raj
·2010-10-02 00:03:54
minimum distance a/√2 and time when thry r closest is a√2v.....
1
sanchit
·2010-10-02 00:40:57
srry but the correct ANS IS min. distance=av1v2 nd min time =av2v
1
varun.tinkle
·2010-10-02 06:00:00
if the distance between 2 moving particles is a at any time then how cant there be the point of closest approach.... think abt this u hv mistyped this ques