Question

a particle moving rectilenearly with constant acceleration is having initial velocity of 10M/S.find out velocity of the particle at the mid point in its path

Answer 1

Answer:

Explanation:

Complete Question,

A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of its path?

Solution,

Let us assume the distance be 2x.

And the distance up to mid point be x.

The, the acceleration be a.

And the velocity be v.

Now,

According to the Question,

We know 3rd equation of motion,

v² = u² + 2as

Putting this, we get

⇒ v² = 10² + 2ax ..... (i)

⇒ 30² = v² + 2ax .....(ii)

Now, Subtracting eq (ii) from (i), we get

⇒ 30² - v² = v² - 10²

⇒ 900 - v² = v² - 100

⇒ 900 + 100 = 2v²

⇒ 1000 = 2v²

⇒ 1000/2 = v²

⇒ 500 = v²

⇒ √500 = v²

⇒ v = 10√5 m/s

Hence, the velocity of the particle at the mid point in its path is 10√5 m/s.

Answer 2

CORRECT QUESTION:-

A particle moving rectilenearly with constant acceleration and having initial velocity of 10m/s . After some time it becomes 30m/s . Find out the velocity of the particle at the mid point in it's path.

GIVEN:

• Particle moving rectilenearly
• With constant acceleration
• Initial velocity = 10m/s

TO FIND:

• Velocity of the particle at mid point

SOLUTION:

Using the third equation of motion

v² - u² = 2as

Where

• v : final velocity
• u : initial velocity
• a : acceleration
• s : distance

Let distance be x

v² = 10² + 2ax

Assuming as equation 1

30² = v² + 2ax

Assuming as equation 2

Subtracting equation 1 - 2

→ 30² - v² = v² + 2ax - ( 10² + 2ax )

→ 900 - v² = v² - 100

→ 2v² = 1000

→ v = √500

→ v = 10√5m/s ( Answer )