Question

A body travelling along a straight line with a uniform acceleration has velocities 5 m/s at a point A and 15 m/s at a point B respectively. If M is the mid point of AB, then choose incorrect statement.

• A. The ratio of times taken by the body to cover distance MB and AM is $\left[\frac{\sqrt{5}-1}{2}\right]$
• B. The velocity at M is $5\sqrt{5}$ m/s
• C. Average velocity over AM is $\frac{5(\sqrt{5}+1)}{2}$ m/s
• D. The product of the acceleration and the distance AB is 200 m²/s².

Answer 1

Answer: (D)

The product of the acceleration and the distance AB is 200 m²/s².

Answer 2

Let $v_A = 5$ m/s and $v_B = 15$ m/s be the velocities at points A and B. Let $d$ be the distance AB. Since M is the midpoint, $AM = MB = d/2$. Using the kinematic equation $v^2 = u^2 + 2as$ for the entire distance AB: $v_B^2 = v_A^2 + 2a(AB)$ $15^2 = 5^2 + 2ad$ $225 = 25 + 2ad$ $200 = 2ad \implies ad = 100$ m²/s². Statement (4) claims $ad = 200$ m²/s², which is incorrect.

To check other statements: Velocity at M ($v_M$): $v_M^2 = v_A^2 + 2a(AM) = 5^2 + 2a(d/2) = 25 + ad = 25 + 100 = 125$. $v_M = \sqrt{125} = 5\sqrt{5}$ m/s. Statement (2) is correct.

Average velocity over AM: $v_{avg, AM} = \frac{v_A + v_M}{2} = \frac{5 + 5\sqrt{5}}{2} = \frac{5(1+\sqrt{5})}{2}$ m/s. Statement (3) is correct.

Ratio of times: Using $v = u + at$: $a t_{AM} = v_M - v_A = 5\sqrt{5} - 5 = 5(\sqrt{5}-1)$. $a t_{MB} = v_B - v_M = 15 - 5\sqrt{5} = 5(3-\sqrt{5})$. $\frac{t_{MB}}{t_{AM}} = \frac{5(3-\sqrt{5})}{5(\sqrt{5}-1)} = \frac{3-\sqrt{5}}{\sqrt{5}-1} = \frac{(3-\sqrt{5})(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{3\sqrt{5}+3-5-\sqrt{5}}{5-1} = \frac{2\sqrt{5}-2}{4} = \frac{\sqrt{5}-1}{2}$. Statement (1) is correct.

Thus, statement (4) is the incorrect one.