Question

A point moving with constant acceleration from $A$ to $B$ in the straight line $AB$ has velocities $v_o$ and $v$ at $A$ and $B$ respectively. Find the velocity at $C$, the mid point of $AB$. Also show that if the time from $A$ to $C$ is twice that from $C$ to $B$ then $v=7v_o$.

Answer 1

According to the second equation of kinematics, we have

$u^2=v^2+2as$

d $=\frac{u^2-v^2}{2a}$ ...........( 1 )

if x is the velocity of point at teh mind point c, then we have

$x^2=v^2+2a\frac{d}{2}$

$d=\frac{x^2-v^2}{a}$ ..............( 2 )

Equation ( 1 ) and ( 2 ), we get

$x^2=\frac{u^2+v^2}{2}$ ...............( 3 )

Hence, the velocity of point at the mind point C is

$x=\sqrt{\frac{u^2+v^2}{2}}$

Now, using first equation of kinematics, we have for

motion between A to C

x = v + $a{t}_1$

$t_1=\frac{x-v}{a}$

for motion between C and B

u = x = $a{t}_2$

$t_2=\frac{u-x}{a}$

since, $t_1=2t_2$

x - v = 2u - 2x

3x = v + 2u

squaring both sides, we get

$9x^2=v^2+4uv+4u^2$ ............( 4 )

substituting equation ( 3 ) in ( 4 ), we get

$u^2+7v^2-8uv$ = 0

Dividing throughout by $u^2$, we get

$7\frac{v^2}{u^2}-8\frac{v}{u}+1$= 0

Hence, we have

$\frac{v}{u}=\frac{8_{-}^{+}\sqrt{64-28}}{14}$ $=\frac{8_{-}^{+}6}{14}$

Hence, $\frac{v}{u}=1$ or $\frac{1}{7}$

Now, $\frac{v}{u}=1$ is not possible as it is a constantly

accelerationg motion.

Hence, $\frac{v}{u}=\frac{1}{7}$

or u = 7v