A body moving with uniform acceleration has v | vedclass

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Question

(A) Let the velocity at point $A$ be $v_A = 20 \, m/s$ and at point $B$ be $v_B = 30 \, m/s$.Let the distance between $A$ and $B$ be $s$. The velocity

Vedclass · Accepted Answer

$25.5 \, m/s$

Vedclass · Answer

(A) Let the velocity at point $A$ be $v_A = 20 \, m/s$ and at point $B$ be $v_B = 30 \, m/s$.Let the distance between $A$ and $B$ be $s$. The velocity $v$ at the midpoint $M$ can be found using the third equation of motion: $v^2 = u^2 + 2as$.For the segment $AM$,$v_M^2 = v_A^2 + 2a(s/2) = v_A^2 + as$.For the segment $MB$,$v_B^2 = v_M^2 + 2a(s/2) = v_M^2 + as$.From the first equation,$as = v_M^2 - v_A^2$. Substituting this into the second equation:$v_B^2 = v_M^2 + (v_M^2 - v_A^2) = 2v_M^2 - v_A^2$.Rearranging for $v_M$:$2v_M^2 = v_A^2 + v_B^2 \implies v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$.Substituting the values: $v_M = \sqrt{\frac{20^2 + 30^2}{2}} = \sqrt{\frac{400 + 900}{2}} = \sqrt{\frac{1300}{2}} = \sqrt{650} = \sqrt{25 	imes 26} = 5 \sqrt{26} \approx 25.49 \, m/s$.Wait,calculating $\sqrt{650} \approx 25.495$. Let's re-check the options. $10\sqrt{6} = \sqrt{600} \approx 24.49$. Actually,the formula for velocity at the midpoint is $v_M = \sqrt{\frac{v_A^2 + v_B^2}{2}}$. Calculation: $\sqrt{650} \approx 25.5 \, m/s$.

$A$ body moving with uniform acceleration has velocities $20 \, m/s$ and $30 \, m/s$ when passing two points $A$ and $B$. Then the velocity midway between $A$ and $B$ is:

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