Question

A particle initially at x=10m, starts moving along the positive x-axis with an initial velocity of 40m/s under the influence of an
acceleration 10m/s directed along the negative x direction. Based on this information, answer the following questions
(i) The particle reverses it's direction of motion at time t0 (t knot) from the start then
(ii) the maximum x co-ordinate of the particles is

Answer 1

Answer:

Ok let's Start ⭐

Explanation:

Note :

★ Newton's 1st equation of motion :

v = u + at

★ Newton's 2nd equation of motion :

s = ut + ½at²

★ Newton's 3rd equation of mother :

v² = u² + 2as

Solution :

★ Given :

• Initial position on x-axis ; x = 10 m

• Initial velocity ; u = 40 m/s

• Acceleration ; a = -10 m/s²

• Final velocity ; v = 0 m/s

★ To find :

• Final position on x-axis = ?

Here ,

The particle is moving in postive x-axis and has negative acceleration which means the velocity of particle is slowing down and gradually the particle will come at rest .

Thus ,

Final velocity , v = 0 m/s

Now ,

Using the Newton's 3rd equation of motion , we have ;

=> v² = u² + 2as

=> 0² = 40² + 2×(-10)s

=> 0 = 160 - 20s

=> 20s = 160

=> s = 160/20

=> s = 80

Hence ,

The distance travelled by the particle till it comes at rest is 80 m .

Thus ,

Final position of the particle on the x-axis = 10 m + 80 m

= 90 m