Question

The points A,B & C lie on the same straight line,the distance AB being 60m.A particle crosses the point A with velocity 2m/s and moves with uniform accerleration along the straight line.After crossing the points B,it takes 10 sec to reach C with velocity 5m/s.Find the accerleration of the particle.

Answer 1

in the given question it is given to you velocity of particle at point A is 2m/s .distance between AB is 60m .

assume velocity of particle at point B is Vb and particle moving with acceleration a.

use velocity-position relationship for path AB

this will be equation 1

For line BC it is given to you it take 3sec to reach point C and also known velocity at point C

apply velocity time relation for path BC.

it will be equation 2.

from equation one and two you can easily calculate acceleration of particle.

Thank you.

Answer 2

The Acceleration Of The Particle Can Either Be 2.1 m/s² Or 0.1 m/s²

GIVEN

A, B, and C lie in a straight line.

AB = 60m

Initial velocity = 2m/s

Time is taken from B to C = 10sec.

Final velocity = 5m/s

TO FIND

Acceleration of the particle.

SOLUTION

We can simply solve the problem as follows-

For distance from A to B

Let the initial velocity be at point 'A' be, Va = 2m/s

Let the Velocity at point 'B' be, Vb

Distance from A to B = 60m

Applying the third equation of motion for A to B

${Vb}^{2} = {Va}^{2} + 2as$

Putting the values in the above formula, we get -

Vb² = 2² + 2 × a × 60

Vb² = 4 + 120a (equation 1)

For Line segment BC

Initial velocity = Vb

Let, Final velocity at point C be, Vc = 5 m/s

Applying the first equation of motion

V = u + at

Putting the values in the above equation we get,

5 = Vb+ 10a

Vb = 5-10a (equation 2)

Putting the value of Vb from equation 2 in equation 1 we get,

(5-10a)² = 4 + 120a

Applying the formula of linear equation

(a+b)² = a² + b² + 2ab

So,

25- 100a+100a² = 4 + 120a

100a²-220a+21 = 0

Solving for we get,

a = 21/10 = 2.1 m/s²

a = 1/10 = 0.1 m/s²

Hence, the acceleration of the particle can either be 2.1 m/s² or 0.1 m/s²

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