| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 May 2008 21:27:42 IST
|
|
|
A car moving on a straight road with a speed 20 m/s. At t=0 the driver of the car applies the brakes after watching an obstacle 150 m. ahead. After the application of brakes the car retards with 2m/s2. Find the position of the car from the obstacle at t=15s.
|
|
|
|
29 May 2008 21:30:56 IST
|
|
|
just give me the answer first....
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:32:22 IST
|
|
|
edited
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:33:10 IST
|
|
|
Is it 50m behind the obstacle?
|
MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
this reply: 2 points
(with 0
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:33:18 IST
|
|
|
how did u get that? i was getting 100....
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:34:36 IST
|
|
|
but wat is the correct ans wer .......... dya knw it???????????
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
The car comes to rest after time given by 0=20-2t t=10s
Distance travelled in these 10s=v2/2a=(20)2/2(2)=100m.
Thus,it is at rest at a distance 50m behind obstacle at t=10s and will be there till t=15s.
|
MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC. |
this reply: 12 points
(with 2
in 3 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:38:05 IST
|
|
|
50 m behind the obstacle
|
_________________________
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 May 2008 21:44:42 IST
|
|
|
ya 50 m behind the obstacle...
lots of answer has already posted...
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
      
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
if i helped u plzzzzz rate me,,,,,,, |
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 13:23:02 IST
|
|
|
well id ont understand how is it 50m
u = 20
t=15
a=-2
so
s= ut + 1/2 at^2
= 20 *15 + 1/2 (-2) * 15*15
= 20 *15 - 15*15
= 15 ( 20-15)
= 15 * 5
= 75 m
so it is 75 m before the obstacle
correct me if i'm wrong
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 13:42:29 IST
|
|
|
car dosent travel 4 15 secons, it comes to rest after travelling 4 10seconds
|
SHREYA |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 14:08:55 IST
|
|
|
4m v=u+at
0=20-2t
t=10sec
S=ut+1/2at*2
so S=100mts
it travelld 100 mts so car is at 50 mts frm d obstacle
|
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 16:59:35 IST
|
|
|
the answer is 75 metres from the obstacle
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 17:01:56 IST
|
|
|
it travels 100m in the 1st 10s and comes 2 rest. but it still has the retardation of 2.hence it travels back 25 metres in the next 5sec.
100-25 is 75.
if wrong plz correct me
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
30 May 2008 22:33:28 IST
|
|
|
look after 10 sec car is at rest and will remain there till 15 sec so ans is 75 m . u are saying car has retardation so it will travel back ,
but normally in kinematics Q we don't consider such cases..
in all problems we assume that after coming to rest body will not be affected by any retardation . otherwise the ans of each Q of each book will change.
|
I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
