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
Explanation:
There's not really enough information to do the problem.
Let's take the common ratio as
We'll call the acceleration
We can arbitrarily set
At
At
The
Given
(AB)/3 = (BC)/2
whereA,B,C are points in sequence on highway
=>(AB)/ (BC) = 3/2
Let
=>AB = 3k, BC=2k
Let constant acceleration
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
2000-2v_A^2>=0
=>v_A<=sqrt1000