Question

A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity $30\,km{h^{ - 1}}$ and $40\,km{h^{ - 1}}$ respectively. The velocity of the car midway between P and Q is?

Answer 1

Hint: Assuming two points P and Q. Let the distance between P and Q be $x$.Velocity os the car is $30\,km{h^{ - 1}}$ and $40\,km{h^{ - 1}}$ .Using the rectilinear formula we can calcite the acceleration of the car using distance as $x$.similarly using that same formula for midway we can calculate the velocity of the car at the midway of P and Q .

Complete step by step answer:
As per the problem a car is moving along a straight road with a uniform acceleration which means the accleartin value of the car remains the same at every point from P to Q. But the velocity of the car varies from point to point. At point P the velocity of the car is $30\,km{h^{ - 1}}$ and at Q the velocity of the car is $40\,km{h^{ - 1}}$. We know from rectilinear formula
${v^2} - {u^2} = 2as \ldots \ldots \left( 1 \right)$
Where, Final velocity of the car = $v$
Initial velocity of the car = $u$
Acceleration of the car = $a$
Distance = $s$

Now from the problem,
$v = 40\,km{h^{ - 1}}$
$ \Rightarrow u = 30\,km{h^{ - 1}}$
$ \Rightarrow s = x$
Putting these values in equation $\left( 1 \right)$ , we get
${\left( {40} \right)^2} - {\left( {30} \right)^2} = 2ax$
$ \Rightarrow 1600 - 900 = 2ax$
$ \Rightarrow 700 = 2ax$
Rearranging the above equation,
$ \Rightarrow a = \dfrac{{700}}{{2x}}$
$ \Rightarrow a = \dfrac{{350}}{x}$
This above acceleration will remain constant for every point from P to Q.

Now applying the same rectilinear formula for mid way we get,
${v^2} - {u^2} = 2as$
Where, $v = {v_m}$, $u = 30\,km{h^{ - 1}}$, $a = \dfrac{{350}}{x}$ and $s = \dfrac{x}{2}$.
Putting these values in the rectilinear equation we get,
${v_m} - {\left( {30} \right)^2} = 2\left( {\dfrac{{350}}{x}} \right)\left( {\dfrac{x}{2}} \right)$
$ \Rightarrow {v_m}^2 - 900 = 2\left( {\dfrac{{350}}{x}} \right)\left( {\dfrac{x}{2}} \right)$
Cancelling the common terms we get,
$ \Rightarrow {v_m}^2 - 900 = 350$
$ \Rightarrow {v_m}^2 = 350 + 900$
$ \Rightarrow {v_m}^2 = 1250$
$ \Rightarrow {v_m} = \sqrt {1250} $
$ \therefore {v_m} = 35.36\,m{s^{ - 1}}$

Hence the velocity of the car midway between P and Q is $35.36\,m{s^{ - 1}}$.

Note: There are three rectilinear formulas. First we have to analyse the problem then according to our requirement we need to choose any of them to find our solution.Remember here the acceleration value should remain constant as the car is moving in uniform acceleration.