Question

Plz Kinematics help:)

Answer 1

Answer is: -\[\color{red}{\text{ Distance(s')= 25m. from the traffic light.}}\]

Answer 2

@ramkrishna Plz help sir:)

Answer 3

\[s=ut+\frac{1}{2}at^2\]

here u=54Km/hr=15m/sec
t=1min=60sec
a=-0.3m/s^2

put these in the above equation and then you will get s=360m

Answer 4

ya i also gt that but answer is 25m. from traffic light:(

Answer 5

distance from traffic light is 40m.

Answer 6

but i have a solution too.
wait a min.

Answer 7

so i have problem
becoz i m getting the same as u i.e.; 40m.
but solution too doesn't seem to be wrong:/

Answer 8

It is the question from this book

Answer 9

sorry, I have to go now. so, i will give my feedback today night.

But as i saw there is some mistake in the solution given by you but can't explain it now.

Answer 10

k. np:)

Answer 11

Yeah 40 meters from the traffic light should be the answer according to the given data, as ramkrishna sir says too.

Where are you being troubled? A mis-print?

Answer 12

How the solution gt t=50sec?

Answer 13

in the attachment

Answer 14

t = 60 seconds actually. since the car's position after 1 minute is being asked, and a minute, my friend if you may remember, is exactly sixty seconds! :D

Answer 15

k:)

Answer 16

thanx a lot:)

Answer 17

No worries, and stop saying 'thanks' every time :P

Answer 18

:D

Answer 19

@maheshmeghwal9 Solution given by you is correct. I find out my mistake.

Given: initial velocity, U=54Km/hr =15m/sec
acceleration, a=-0.3m/s^2
We have to find out the position of car at t=1min=60sec.


NOTE: SINCE MY CAR IS DEACCELERATING SO A TIME COMES WHEN MY CAR IS STOP. AFTER THAT TIME CAR WILL NOT MOVE. LET SAY THAT TIME IS " t ", AND AT THAT TIME THE POSITION IS " X". SO, THE POSITION OF CAR AT ANY TIME " t + dt " IS SAME " X " BECAUSE THE CAR IS NOT MOVING.


STEP: 1 : Lets first find out the time at which car is at rest. At this time final velocity, V=0m/s.

So, using equation

\[V = U + at\]
\[0 = 15-0.3 t\]
\[t=\frac{15}{0.3}=50\sec\]

Since at t =50 second car comes to rest so at any time greater than 50 second car will always be at the same position as at time t=50sec.

Thus,
\[V^2-U^2=2as\]

here at t=50 sec V=0m/sec, U=15m/sec, a=-0.3m/s^2

On substituting these we get s=375m.

So, car will be at a distance 25cm from the traffic light.

Answer 20

oh i see. thanx a lot sir

Answer 21

but sir maine aisa sawal aaj tak nahin dekha:/

Answer 22

Don't tell a lie. Aj tum subha se ye sawal dekh rahe ho....... :D

Answer 23

lol

Answer 24

btw sir aapka OS mein aane ka timetable kya hai?

Answer 25

in a period of 5 to 6 days ????????????

Answer 26

OS???

Answer 27

Open Study.

Answer 28

koi fix time nahi hai. But generally mein rat mein 10 baje ke as-pas ata hun.

Answer 29

agar koi question ho to tag kar diya karna...

Answer 30

oh i see thanx a lot for info
woh kya hai ki main IIT kar raha hoon
isliye khass apki madad chahata hoon:)
haan koi question hoga to main aapko jaroor tag karung :D

Answer 31

karunga*

Answer 32

IIT-JEE ki preparation ke liye best of luck. But you have to work too hard because first 5-6 IITs mein hi admission milega to hi thik hoga, kunki nai IITs thik nahi hai....

Answer 33

k! sir aapki baat hamesha yaad rakhoonga:D

Answer 34

thanx for ur wish:)

Answer 35

waise sir aapko main ek free physics books ki site doon?
usmein aap something 4000 books free payenge
maine bhi kaphi books download ki hain:)

Answer 36

if u want then ?

Answer 37

GREAT!! I should have thought about that!

Answer 38

hmm..........:)

Answer 39

Ha de do............

Answer 40

k! sir:)

Answer 41

http://freebookspot.es

Answer 42

but remember no need to log in:)