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: (B), (C)

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. Let $a$ be the acceleration and $d$ be the distance AB. From $v_B^2 = v_A^2 + 2ad$, we get $15^2 = 5^2 + 2ad$, which simplifies to $225 = 25 + 2ad$, so $2ad = 200$, and $ad = 100$. Statement (4) is incorrect.

For statement (2), let $v_M$ be the velocity at M. $v_M^2 = v_A^2 + 2a(d/2) = v_A^2 + ad = 5^2 + 100 = 125$. So $v_M = \sqrt{125} = 5\sqrt{5}$ m/s. Statement (2) is correct.

For statement (3), average velocity over AM is $\frac{v_A + v_M}{2} = \frac{5 + 5\sqrt{5}}{2} = \frac{5(\sqrt{5}+1)}{2}$ m/s. Statement (3) is correct.

For statement (1), time taken for AM is $t_{AM} = \frac{v_M - v_A}{a} = \frac{5\sqrt{5} - 5}{a}$. Time taken for MB is $t_{MB} = \frac{v_B - v_M}{a} = \frac{15 - 5\sqrt{5}}{a}$. The ratio $\frac{t_{MB}}{t_{AM}} = \frac{15 - 5\sqrt{5}}{5\sqrt{5} - 5} = \frac{3 - \sqrt{5}}{\sqrt{5} - 1} = \frac{(\sqrt{5}-1)}{2}$. Statement (1) is correct.