Question

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?

Answer 1

`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 }`

Answer 2

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`