Saveetha Engineering College, Chennai All Questions

Hello dear student !

Please elaborate your question more and rewrite it with more explanation so that we can answer your query more properly.

Dear aspirant

Average acceleration is  change in velocity upon change in time.

Therefore

Acceleration= v÷t

=20÷5

=4   will be average acceleration.of car as per your question.

Thank you

Hello student yes you can but that's totally depend on the administration department of both of the college first you need to ask the college where you want to take admission if there's a vacant seat available or not in that particular branch if they agree to avail you the admission after that you need to ask/request your college that you want to take admission in a different college so please grant transfer certificate for that.
When both the colleges mutually agree for the same which happens very rarely in practical cases then you can take admission for sure.

Feel free to comment if you have any doubt
Good luck

Hello,

First get your concepts clear, you can start off with Grammar, learn all the important topics such as Subject-Verb Agreement, Preposition, Noun, Pronoun, Adjective, Adverb, Active & Passive Voice, Narration etc. You can go through Wren and Martin for this, Having a strong foundation on grammmar will enable you to solve sentence reaarangement, fill in the blanks, error detection etc,

You can work on your vocabulary by reading any newspaper or going through Word Power Made Easy, it's a wonderful book for vocabulary.

Once you have gone through aforementioned steps, then try solving any English Book, you can choose between Plinth to Paramount or SP Bakshi. Along with this, go through previous year papers, anaylse on a regular basis.

Hello,

After diploma in EEE,  the students can choose to pursue degree of B. Tech in engineering colleges. The engineering colleges take admission of the diploma students through lateral entry. This can help them to study about the advanced technologies being taught to the regular B. Tech students. It can also help them to make an attractive resume for placement sessions.

Hello,

We know that, as per the Equation of Motion,

s = ut + 0.5 a t^2

So, for first 3 sec.,

10 = 3u + 0.5 x a x 9

So, 3u + 4.5a = 10.............. ( 1 )

Now, for next 4 sec.,

25 = 7u + 0.5 x a x 49

So, 7u + 24.5a = 25............ ( 2 )

Solving ( 1 ) and ( 2 ), we get,

a = 0.12 m/s^2 and u = 3.15 m/s

Now, also as per the Equation of motion,

v = u + at

So, For 8 sec.,

v = ( 3.15 ) + ( 0.12 x 8 )

So, v = 4.11 m/s

Hence, The velocity of object at the end of 8 seconds will be 4.11 m/s

Best Wishes.

Hello,

Given that the distance is directly proportional to the square of time.

So, d ∝ t^2

So, d = k.t^2

Where, k is the constant of proportionality.

Now from first condition,

d = 78m and t = 7 sec.

So, 78 = k. 7^2

So, k = 1.59 = 1.6

Now, d = k. t^2

So, for first four seconds, t = 4sec

So, d = k. 4^2

So, d = 1.6 x 16

So, d = 25.6 m

So, The stone will fall a total distance of 25.6 m in 1st 4 seconds.

Best Wishes.

Hey ..
I think the formula for velocity of a particle in SHM is
Velocity = omega ( amplitude - displacement )
So the answer for the given terms would be
V = w (a-x)
w= 2/T.     where T is time period
w= 2 rad /sec
a = 5 cm
x = 3 cm
So V = 8 cm/sec
Hope the answer is right...
All the best..

Hello poonam
It is good to hear that your daughter is taking part in olympiads from such a little age. There are some factors which affect the rank in these olympiads, one of them is alphabetic order of the name. Also, sometimes there is some mistake in the omr that gets neglected. I'd say don't take stress because she's at a very tender age and you should just motivate her instead of comparing her with other students.

Using s = ut + 1/2a(t^2),  we can write

100 = 4u  + 8a

=> u + 2a = 25             ------(1)

Also, Since the distance traveled in the next 4 seconds is 140m, the total distance traveled in the 8 seconds is 240 m. So we can write

240 = 8u + 32a

=> u +4a = 30              ------(2)

Solving Eq.1 and 2 we can say,

a = 5/2 m/s/s and u = 20 m/s

Hence the velocity at t=6 sec, is 20 + 5/2*6 = 35m/s.