A,B,C are points in sequence on highway such that AB/3 = BC/2. Find velocity at A & C, at B it is 20m/s. If acceleration is constant?

2 Answers
May 25, 2018

v_a = - sqrt{v_b^2- 6 r g }

v_c = sqrt{v_b^2+ 4 r g }

Explanation:

There's not really enough information to do the problem.

Let's take the common ratio as r so AB=3r, BC=2r.

We'll call the acceleration g and the given velocity v_b=20 m/s.

v = v_0 + g t

x = x_0 + v_0 t + 1/2 g t^2

We can arbitrarily set t=0 to be when we pass B and set x_0 = x_b = 0.

v = v_b + g t

x = v_b t + 1/ 2 g t^2

At x=-3r we get

-3r = v_b t_a + 1/ 2 g t_a ^2

g t_a^2 + 2v_b t_a + 6 r = 0

t_a = 1/g text{} (-v_b pm sqrt{v_b^2- 6 r g } )

v_a = v_b + g t_a = pm sqrt{v_b^2- 6 r g }

At x=2r we get

2r = v_b t_c + 1/2 g t_c^2

0= g t_c^2 + 2v_b t_c - 4r

t_c = 1/g text{} (-v_b pm sqrt{v_b^2 + 4rg})

v_c = v_b + g t_c = pm sqrt{v_b^2 + 4rg}

The pm are telling us this is just like a ball thrown up in the air. If we call the time and height zero at the apex, the motion is symmetrical; at a given height we'll be going the same speed down as up. From the problem statement we can choose the negative time for t_a and the positive time for t_c.

v_a = - sqrt{v_b^2- 6 r g }

v_c = sqrt{v_b^2+ 4 r g }

May 25, 2018

Given (AB)/3 = (BC)/2
where A,B,C are points in sequence on highway
=>(AB)/ (BC) = 3/2

Let k be common factor of the ratio

=>AB = 3k, BC=2k

Let constant acceleration =a, Let velocity at point A be=v_A and velocity at point C be=v_C

Using the kinematic expression

v^2-u^2=2as

we get following two equations

(20)^2-v_A^2=2a(3k) ......(1)
v_C^2-(20)^2=2a(2k) ......(2)

Dividing (1) by (2) we get

(400-v_A^2)/(v_C^2-400)=3/2
=>800-2v_A^2=3v_C^2-1200
=>3v_C^2=2000-2v_A^2
=>v_C=sqrt((2000-2v_A^2)/3)

For v_C to be real,

2000-2v_A^2>=0
=>v_A<=sqrt1000