Question

solve this problem pls no spam​

Answer 1

Explanation:

Hope you are enjoying kinematics xD It can enjoyed but the problem is that the questions from this chapter especially these type makes you think a lot making you feel stressed and especially if you choose to tackle these questions without drawing the figures.. I really have no clue how much time you will have to spend on a particular question,so always imagine or better draw the diagram according to the question and proceed then.

So,the first question is not that tough,the other two aren't tough as well but you can solve them only if you have sufficient practice.

Let's begin.

Solution 1 :

In the first question we actually have been given three different phases of the motion of the particle.

First phase → when the particle starts from rest possessing an acceleration of 2 m/s² for t = 10 seconds

Second phase → It travels with const speed for t = 30 s

Third phase → The particle undergoes const retardation of 4 m/s² and comes to rest.

So,in all these three phases the particle must have covered some distance, let's call them S1 for first phase,S2 for second phase and S3 for the third phase.

Now let's calculate these distances and we will add them all to get the final distance covered by the particle.

S1 = ut + ½at²

S1 = ½ × 2 × 10² [ ∵ u = 0,ut becomes zero]

S1 = ½ × 2 × 100

S1 = 200/2

S1 = 100 m ---> (1)

Now before we calculate S2,we need to calculate v for this case. So, we can easily do it.

v = u + at

v = 0 + 2 × 10

v = 20 m/s

Now,we have everything ready to calculate S2. Since mentioned in the question that the particle moves with constant speed,we can directly use the formula S = vt to find the distance in the second phase, given time,t = 30 s

S2 = vt

S2 = 20 × 30

S2 = 600 m ---->(2)

Now,all we need is S3, let's calculate.

In this phase,v = 0 m/s (because eventually the particle is going to come to rest),a = 4 m/s²,u = 20 m/s

[Fact : The intial velocity of the next phase of motion is same as the same final velocity of the previous phase, hence the final velocity v = 20 m/s in second phase is the intial velocity,u = 20 m/s in the third phase].

We know,

v² = u² + 2as

0² = 20² + 2 × 4 × S3

(-20)² = 8S3

400 = 8S3

400/8 = S3

S3 = 50 m ---->(2)

Now add (1),(2) and (3),

S = S1 + S2 + S3

S = 100 + 600 + 50

S = 750 m

So, option (C) is correct.

Now moving onto the next question.

Solution 2 :

Here as well we are going to follow the same logic of dividing the journey of particle in different phases.

First phase → u = 0 m/s,acceleration = α (const),t = t1

Second phase → acceleration = β (const),t = t2

Let's first move to the first phase and calculate the final velocity,v.

v = u + at

v = 0 + αt1

v = αt1 -----> (1)

Now,the distance covered in this phase we need to calculate.

Now moving to the second phase.

v = u + at

v = at

v = βt2 ----> (3)

Comparing (1) and (3), we get :

αt1 = βt2

t1/t2 = β/α ---->(4)

Now we will calculate S1 and S2.

v² = u² + 2aS1

v² = 2aS1

S1 = v²/2a

S1 = v²/2α (a = α in first case)

Now let's calculate S2.

v² = u² + 2aS2

0 = v² + 2βS2 (u = v, simplification purpose)

-v² = 2βS2

S2 = -v²/2β

The total distance is given in the question as S.

So,

S = S1 + S2

S = v²/2α - v²/2β

S = v² [ (1/2α) - (1/2β)

S = v² [ 2α - 2β/ 4αβ]

(4αβ)S = v² (2α -2β)

(4αβ)S/(2α - 2β) = v²

v² = 2 (2αβ)S / 2( α - β)

v² = 2αβ S/α - β

v = [2αβ S/α - β]^(½)

Option (D) is the correct choice.

Now let's do the last question.

Following the same logic here as well, dividing the journey into phases.

First phase : Distance from the fixed point is a,t = 0

Second phase : Distance from the same point is b,t = n seconds

Third phase : Distance from the same point is now c,t =2n seconds.

Let's move to the first phase and calculate the distance particle covered in n seconds.

So the particle starts from the intial point at t = 0 and reaches point b in n seconds and also given it is at a distance a from the point at t = 0 s, so distance covered in n seconds = b - a.

Now we will calculate the acceleration (a').

S' = ut + ½a't²

(b -a) = u (n) + ½ a'(n²)

Now taking LCM,

2b - 2a = 2un + a'n² ---- (1)

Now let's move to the second phase, distance covered in this phase will be the difference of the particle's position at 2n second and at t = 0 i.e a.

Mathematically : Distance (S") = c - a

Acceleration in this case,

(c - a) = u (2n) + ½a'(2n)²

c - a = 2n • u + ½ × 4n² × a'

c - a = 2n • u + 2n² × a' ----> (2)

Now subtract equation (2) from (1),

c - a - (2b - 2a) = 2nu + 2n²a' -(2nu + a'n²

c - a - 2b + 2a = 2nu + 2n²a - 2nu - a'n²

c - 2b = 2nu - 2nu + 2n²a' - a'n²

c - 2b + a = a'n²

c - 2b + a /n² = a'

So,the acceleration,a' = c - 2b + a/n²

Option (A)

(Diagram for the third question attached for convenience)

PS : Sorry if I forgot to explain certain points, I was getting impatient xD had no more capacity to type anything else *sighs*

Answer 2

Answer:

Find out the word in the passage with the same meaning

Explanation:

hello hello

:)

noi me khussi