Question

A point moving with constant acceleration from A to B in a straight line AB has velocities u and v at A and B respectively, 'C'is midpoint of AB. Find the ratio
if time taken from A to C is twice the time to go from C to B.
A
1
3
5
7
​

Answer 1

Let us assume that the total distance covered is x km. (From A to B distance covered is x km).

Now, using the Third Equation Of Motion. We can say that,

u² - v² = 2ax (s = x)

(u² - v²)/2a = x ..................(1st equation)

If 'C' is mid point between AB then AC = x/2 and CD = x/2

Assume that velocity at point C is y.

So, distance is:

y² - v² = 2a(x/2)

y² - v² = ax

(y² - v²)/a = x ...................(2nd equation)

On comparing 1st & 2nd equation, we get

(u² - v²)/2a = (y² - v²)/a

u² - v² = 2y² - 2v²

(u² + v²)/2 = y² ..............(A)

√[(u² + v²)/2] = y ...............(3rd equation)

Also assume that the time taken between A to C is t. As said in question time taken from A to C is twice the time to go from C to B. So, time taken from C to B is 2t.

Using first equation of motion, (Form A to C)

y = v + at

(y - v)/a = t ...............(4th equation)

From C to B

u = y + at

(u - y)/a = t .............(5th equation)

As per given condition,

(y - v)/a = 2(u - y)/a

y - v = 2u - 2y

3y = 2u + v

Squaring both sides,

9y² = 4u² + v² + 4uv

Substitute value of (A) above

9u² + 9v² = 8u² + 2v² + 8uv

u² + 7v² - 8uv = 0

Divide by u²

u²/u² + 7v²/u² - 8uv/u² = 0

1 + 7v²/u² - 8v/u = 0

7v²/u² - 8v/u + 1 = 0

Using quadratic formula,

v/u = [8 ± √{(-8)² - 4(7)(1)}/2(7)]

v/u = [8 ± √(64 - 28)/14]

v/u = (8 ± 6)/14

v/u = 14/14, 2/14

v/u = 1, 1/7

For constant acceleration a can't be 1. So,

v/u = 1/7

7v = u

Answer 2

Answer:

Explanation: see the pics