# 11. Class 11th Physics | MOD | Solved Example-7 on Uniform Acceleration | by Ashish Arora

In this e.g. we are given that a particle

starts, at initial speed of seventeen meters-per-second, and it retards at two meters-per-second-square,

and we are require to find the distance covered by the particle in, ninth second of motion.

We just draw the physical situation in this question. We are having a particle, at t=

zero it starts with an initial speed of, seventeen meters-per-second, and it is retarded, with

an opposite acceleration of two-meter-per-second-square. So gradually it’s speed will decrease. and

after some time it’s final velocity will become zero. And we can easily calculate the

time at which it’s final velocity is becoming zero, say this time is t-one so we can simply

write the equation v-final=u+ (a-t). so here, zero can be written as seventeen minus two-t-one

because acceleration will be minus in this case, so here we can see the time t-one we

are getting is seventeen by two i.e. eight point five second. So at, t-one=eight point

five second it is coming to rest, and then it will return back, because acceleration,

was, opposite to this point , what will happen , it’s velocity is retarded it will come

to rest and then, it returns back due to the continuous action of acceleration and now

it will be in an, accelerated motion . now in this journey we are required to find the

distance covered by the particle in ninth second of motion. We calculate, the distance

by using the formula of displacement of n-eth second, or displacement of particle. In n-eth

second . just see what will happen , it will be able to, have look on interesting result,

the formula for n-eth second of motion was u+, (a) by two into two-n minus one. So here

we are required to analyze , the motion in ninth second . so we find out the displacement

in nine second of motion this will be seventeen , + (a) by two , this will be minus two by

two, into two-into nine minus one, just see what are you getting this seventeen minus

. this is one, and this eighteen minus one is seventeen, the result is zero . now it

is quite a surprising result how it is possible that displacement becomes zero, (yes) it is

possible, because you can see we are required to analyze the motion in nine second , so

nineth second will be the duration from t=eight to t=nine second of motion. This is

the ninth second, so we can see at eight point five second , the particle was at rest, it

is the extreme position, so if it is at this position at t=eight second. It will take

point five second, in going from, this point to this point . and similarly, as we analyze

the accelerated motion with same acceleration it will take same time that is point five

second in going from this point to this point back, so here it will reach at t=nine second

. so you can see at t=eight and nine second particle was at same position that’s why

the result is coming out to be zero. Now this is the formula for. displacement in nine second

it is not given as the distance . and in the question it is (ask) to find the distance

covered, so that we can easily calculated, by finding out the distance covered by the

particle from rest, within point-five second and we make it twice. So, if this distance

is, s–one, this s-one can be directly be written as half (a-t) square , as initial

speed here is zero, so this half into two into zero-point-five whole-square. So this

is zero point two- five meter, so we can directly write distance covered. In ninth second is.

This two-s-one i.e.=zero point five meter which is the answer to the question. So be

careful about such type of questions. Which are slightly tricky, but several time it happen

that students without, any analysis, or without any thinking they just apply, the direct formulas,

and they ignore such type of language in the question . so be careful in future about such

type of questions.

## 12 Replies to “11. Class 11th Physics | MOD | Solved Example-7 on Uniform Acceleration | by Ashish Arora”

why is initial speed 0 during s1

sir i understand the way you solve the problems but when i try to solve problems i am stuck in some or the other point. While talking about questions like this one i cant even try to attempt also. They are really hard.But i undestood this question..Your explaination helps me alot sir. Thanxx

why didnt we find distance between 8 to 8.5 sec ( u = 17 ) , then multiply it twice . it gives different results ??

sir if speed is 0 at 8.5 how can v calculate for t=9? like it has been retarded and come to rest so how?

sir there is some problem with qusets they are not opening and even interact blog is not opening

Sir 8 to 8.5 sec me agar ham distance nikalenge tab to initial velocity bhi rahegi 17 and answer different aa jayega

sir the solution of quset is not given?

time in last why u take 0.5. sec

Distance covered in 9 second is why 2s1 plz any one reply

Sir very nice question clears a lot of concept

Sir why are we not taking -ve sign at 4.00???

Sir the question says find the distance in 9th second but why do you calculate distance in 8th to 9th second particularly??