a boy moving with constant acceleration from A to B in the straight line AB has velocities vA & vB at A & B respectively. Time taken by the boy to move from A to C is twice of that from C to B, where C is mid-point of AB. The ratio of vA to vB is_
Answers
Answer: 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 = .(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
Explanation:
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.(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