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

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”

  1. 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

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

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

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

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