STD 11 - SCI - ENG - PHYSICS - CH : 02 - EXE 21 05

Ajouté : 2026-05-26

[00:00:11]Good morning, students.

[00:00:17]Good morning, students.

[00:00:19]Today we are going to see exercise problem 2.4.

[00:00:24]Here in this problem a drunkard a drunken person walks in a narrow lane.

[00:00:38]When he walks he walks five steps forward and three steps backward.

[00:00:47]Always five steps forward, three steps backward. This way he walks.

[00:00:55]Each step completed in 1 second duration.

[00:00:59]To cover one step it require 1 second time.

[00:01:04]So, always walk five steps five steps forward, three steps backward. Again five steps forward, three steps backward.

[00:01:15]This way he walk and ultimately he fell down into a pit which is 13 m away from his initial position.

[00:01:28]So, pit is at 13 m away from its initial from his initial position.

[00:01:36]So, problem is given like this. A drunkard walking in a narrow lane takes five steps forward and three steps backward followed again by five steps forward and three steps backward and so on.

[00:01:51]Each step is 1 m long and requires 1 second.

[00:01:57]Each step is also 1 meter long and requires 1 second time.

[00:02:02]Plot XT graph of his motion.

[00:02:07]We have to plot graph of position versus time. Each step has a length 1 meter and to complete one step he require 1 second time.

[00:02:16]Determine graphically and otherwise how long the drunkard takes to fall in the pit 13 meters away from the start.

[00:02:26]So, here I have written a drunkard walks one step in 1 second.

[00:02:31]One step is of length 1 meter and it requires duration 1 second.

[00:02:39]He walks always five steps forward and three steps backward.

[00:02:47]Five steps forward, three steps backward.

[00:02:50]This kind of repeated motion he perform.

[00:02:55]So, position versus time graph is given to you.

[00:03:03]Time always plotted on a positive x-axis, while position position is given on a positive y-axis.

[00:03:12]Initially, time T is zero.

[00:03:17]Position is also zero.

[00:03:20]I mark 1 2 3 4 5 up to 40 seconds on the positive x-axis.

[00:03:30]Time is given.

[00:03:32]And on y-axis I mark position 1 2 3 4 5 6 7 8 9 10 11 12 13 like that. Position is given in a meter.

[00:03:48]Position is given in a meter.

[00:03:51]Okay?

[00:03:53]Now, person started journey.

[00:03:55]Drunkard started journey in a narrow lane.

[00:03:59]He started journey from origin at time t is equal to zero.

[00:04:05]Then he made a total like five step forward. So, each step Each step of 1 m long means he walks 5 m.

[00:04:17]He walks 5 m during 5 seconds. Each step required 1 second time. So, 1 2 3 4 and 5.

[00:04:25]So, after 5 seconds his position 5 m away.

[00:04:30]So, I have marked point 5 5.

[00:04:33]I will represent time on x-axis while this 5 represent position on y-axis.

[00:04:41]So, position and time both points given.

[00:04:46]So, at this point again next three step he walk backward.

[00:04:53]So, again position decreases with respect to origin.

[00:04:57]So, you can see from fifth step again three step back backward, he reach at a 2 m.

[00:05:06]He reach at what? 2 m. 5 minus 3 means 2.

[00:05:10]But, time never come back.

[00:05:13]Time move forward.

[00:05:16]So, after 5 seconds each step required 1 second means three step three step required three second time. So, time on time axis 5 after 1 means 6, 6 after 1 7, 7 after 1 8.

[00:05:34]Means drunkard required another 3 seconds to move backward three step.

[00:05:40]And again his position come at 2 m.

[00:05:44]But, time reach at 8 seconds.

[00:05:47]So, 5 + 3 seconds means 8 seconds. So, I mark this point 8, 2 means it represent time 8 seconds while position represent at 2 m.

[00:05:59]From origin with respect to origin, his position position is now 2 m.

[00:06:05]Thereafter, thereafter, he started move forward.

[00:06:11]Forward five step.

[00:06:13]So, from 2 m, if I add 5 m again, so what is my final position? 2 + 5 2 + 5 means 7. So, I will reach at a 7.

[00:06:28]So, again five step forward requires more five second time.

[00:06:33]So, 8 + 5 five second time.

[00:06:37]8 + 5 when you add, 8 + 5 is 13.

[00:06:40]So, his position at a 13 second 13 second. I mark 13 seconds.

[00:06:46]And position 2 + 5 means 7 m.

[00:06:50]So, 13, 7. 13 second is a time, 7 is meter distance.

[00:06:56]Now, at this moment, again he move backward backward of three step.

[00:07:07]Three step means from 7 m, again he walk three step backward. So, he will reach 7 - 3 4 4 m.

[00:07:16]So, here I mark 4 m.

[00:07:19]But time to add After 13 seconds, another three seconds if I add for three seconds each step he walks in a 1 1 second. So, three step require three seconds. So, 13 + 3 time 16 seconds.

[00:07:37]To move backward, he requires time.

[00:07:41]And that time is added to the previous time. So, 13 + 3 seconds gives a 16 second time.

[00:07:48]And position is moving backward towards initial point. So, 7 - 3 that gives a 4 m. So, position is at the 4 m.

[00:07:59]Now, he walks forward.

[00:08:03]He walks forward how much meter?

[00:08:06]He walks five steps forward means 5 m forward.

[00:08:09]So, 16 + 5 That means time 21 seconds.

[00:08:16]Five Five steps requires 5 seconds time.

[00:08:19]So, first on x-axis I have represented time. So, 16 + 5 seconds means 21 seconds.

[00:08:26]And position 4 + 9 4 + 5 means 9.

[00:08:30]4 plus five steps of each of 1 m means 4 + 5 that gives a 9 m. So, I mark the point at the 9 m.

[00:08:40]So, five steps forward 9 m.

[00:08:42]Time 21 seconds. Position is at 9 m.

[00:08:46]Now, again he walk backward.

[00:08:49]So, 21 seconds thereafter another 3 seconds for three steps.

[00:08:56]Though he move backward, but time required 3 seconds that should be added to previous time. So, 21 + 3 so on time axis it comes 24. So, at time 24 seconds his position position initially 9 m with respect to origin. Now, for three steps backward of 3 m each of 1 1 m. So, 9 - 3 that gives a 6.

[00:09:20]So, position is a mark at a 6 m.

[00:09:24]His position is marked at a 6 m.

[00:09:28]6 m position while time is at 24 seconds.

[00:09:33]Now, at this moment further he walk five steps forward, each of 1 m.

[00:09:40]So, time required to complete five steps is a 5 seconds, so 24 + 5 that is a 29 seconds.

[00:09:49]After 29 seconds, his position 6 + 1 6 + 5 each of step 1 1 m long means six five steps forward 6 + 5 that gives 11. So, 6 + 5 m that gives you 11 m to you.

[00:10:08]See, this is 11 m for you.

[00:10:13]11 m.

[00:10:14]The time is 29 seconds.

[00:10:17]Again, further he move backward backward three steps backward as usual.

[00:10:25]Three steps backward means three steps backward three seconds time.

[00:10:29]So, time 29 + 3 seconds added. 29 + 3 seconds means 32 seconds.

[00:10:36]At 32 seconds, but position with respect to origin as he start move backward so from 11 three steps backward means 3 m backward.

[00:10:47]11 - 3 that gives 8 m.

[00:10:52]8 m. His position is marked at 8 m.

[00:10:56]Okay? We can represent this point as 32 seconds and 8 m.

[00:11:04]This gives the time and position.

[00:11:10]On the x-axis we have represented time.

[00:11:13]On y-axis we have represented position.

[00:11:16]Position is x.

[00:11:19]So, it is at 8 m.

[00:11:21]Now, he move forward.

[00:11:26]He move forward of the step five steps.

[00:11:30]So, from 18 m 8 m, from 8 m, further 5 m is added each of 1 1 m step, high step. So, 8 + 5, 8 + 5, 13. So, position is marked at a 13 m.

[00:11:49]Position is marked at a 13 m.

[00:11:52]Time, 5 second time is required.

[00:11:56]32 + 5 second.

[00:11:59]So, this 13 m position marked at 37 second time.

[00:12:06]And our question is given like this.

[00:12:10]From his initial position, origin, at a distance 13 m, there is a pit.

[00:12:19]A person walk forward, backward, this way.

[00:12:23]He fell at a distance 13 m pit.

[00:12:27]So, after 13 m, no need to plot the graph further.

[00:12:33]So, at 13 m, there is a pit in which person fall.

[00:12:37]At what time he fall?

[00:12:39]He fell inside the pit.

[00:12:42]That is 30 second.

[00:12:44]From his start, at time 37 second, he fell into the pit.

[00:12:51]So, T is 37 second, X is 13 m.

[00:12:56]Hence, person will fall into the pit after 37 second.

[00:13:01]So, this is position time graph.

[00:13:04]You can easily understand position time graph of motion of this drunkard person.

[00:13:13]He always walk five step forward, three step backward.

[00:13:18]But, on time axis, he move forward, forward, forward, only forward.

[00:13:23]Position come back, but time never come back.

[00:13:26]So, you might observe this thing.

[00:13:30]You can easily understand this thing.

[00:13:33]Next.

[00:13:51]Exercise 2.5.

[00:13:55]A car moving along a straight line on a straight highway with a speed 126 km/h is brought to a stop within a distance of 200 m.

[00:14:09]What is the retardation of car and how long does it takes for the car to stop?

[00:14:17]Car is driven with speed 126 km/h.

[00:14:23]So, initial velocity 126 km/h.

[00:14:29]And once driver apply brake after apply brake car travel 200 m, then it stop.

[00:14:42]So, braking distance is 200 m.

[00:14:45]Car stop, vehicle stop means its final velocity is zero.

[00:14:51]Now, for this car, retardation is done.

[00:14:57]Retardation means deacceleration.

[00:14:59]Acceleration, but negative because it is retardation, velocity decreases. Once we apply brake, velocity decreases and for decreasing velocity we can use the word retardation or deacceleration.

[00:15:16]Here, braking distance is a 200 m, S or you can say DS, d distance, s stopping distance.

[00:15:24]200 m. Initial velocity of car that is given v 0 126 km/h.

[00:15:31]We need to convert this km/h into m/s.

[00:15:35]In 1 km that is 1,000 m.

[00:15:38]Upon hour, hour has a second 3600.

[00:15:42]3600 seconds. Two zero cancel.

[00:15:46]So, 10 by 36 into 126 calculate.

[00:15:50]126 into 10 by 36.

[00:15:54]Uh by two, two fives are 10.

[00:15:58]Two 18s are 36. So, five by 18.

[00:16:02]Now, 18 sevens are 126.

[00:16:05]So, seven into five.

[00:16:07]Seven into five 35 m/s.

[00:16:10]So, initial velocity of car given to you is a 35 m/s.

[00:16:16]When car stop, its final velocity is zero.

[00:16:19]When car stop, its final velocity v becomes zero.

[00:16:24]Now, to determine acceleration, to determine retardation, to determine deacceleration, we can apply third equation of motion.

[00:16:34]Third equation of motion v squared minus v zero squared that is equal to two a s.

[00:16:39]v final velocity that is zero.

[00:16:43]Minus initial velocity 35 m/s so 35 squared is equal to two a s.

[00:16:51]Braking distance 200 m.

[00:16:54]So, it is 200.

[00:16:56]Zero squared zero. No need to write down that zero. Minus 35 squared 1225.

[00:17:03]That is equal to two into 200 means 400 into a.

[00:17:08]Make the a subject. Minus 1225 by 400.

[00:17:13]So, answer is minus 3.0625.

[00:17:16]So, deacceleration retardation comes negative.

[00:17:21]Minus that represent it is retardation or deacceleration.

[00:17:26]3.0625 m per second squared.

[00:17:30]So, we got the answer of retardation.

[00:17:33]And retardation we got minus 3.0625 m per second squared.

[00:17:40]Now, another question is also asked to you.

[00:17:44]How much time this car take to stop? Once that driver apply brake, immediately car will not stop.

[00:17:51]It travels some distance. So, to travel some distance, it also requires some time.

[00:17:57]So, we are going to derive this time. How much time it require to stop the vehicle?

[00:18:08]Here.

[00:18:10]Time taken to stop the car.

[00:18:13]We can apply first equation of motion. V is equal to V0 plus AT.

[00:18:18]Acceleration we already determined.

[00:18:21]Time is asked to you.

[00:18:23]V final velocity, V0 initial velocity.

[00:18:26]Car V When we apply brake, car ultimately stop means its final velocity is zero.

[00:18:34]Its initial velocity before we apply brake, that is given to you 35 m per second.

[00:18:39]So, easily we can apply equation of motion one.

[00:18:44]Final velocity V is equal to V0 plus AT.

[00:18:46]Final velocity is zero, initial velocity 35, plus acceleration A is minus 3.0625 into time.

[00:18:57]When you take 35 on opposite side, that is minus 35 is equal to minus 3.0625 T.

[00:19:04]Minus minus cancel.

[00:19:05]>> [clears throat] >> 35 upon 3.0625.

[00:19:10]That gives 11.43.

[00:19:13]Time T is That T is 11.43 seconds. It is plus time never a negative.

[00:19:21]So, T is equal to 11.43 seconds.

[00:19:26]That is time required by vehicle to stop once a driver apply brake.