Question

In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of car B. Assuming that both the cars start from rest and travel with constant acceleration ${a_1}$ and ${a_2}$ respectively. So, the value of v will be :
${\text{A}}{\text{. }}\left( {\sqrt {\dfrac{{{a_1}}}{{{a_2}}}} } \right)t \\\ {\text{B}}{\text{. }}\left( {\sqrt {\dfrac{{{a_2}}}{{{a_1}}}} } \right)t \\\ {\text{C}}{\text{. }}\left( {{a_1}\sqrt {{a_2}} } \right)t \\\ {\text{D}}{\text{. }}\left( {\sqrt {{a_1}{a_2}} } \right)t \\\$

Answer 1

Hint: We are given the velocity, acceleration and time taken by the cars A and B and also how they are related to each other. By writing the various equations of motion for A and B, we can get the value of v.

Formula Used:
Equations of motion used in the solution are.

${\text{v}} = u + at \\\ S = ut + \dfrac{1}{2}a{t^2} \\\$
where u is the initial velocity, v is the final velocity, a is acceleration, t is time taken and S is the total distance covered by the moving body.

Complete step-by-step answer:

Let car A takes ${t_1}$ time to travel from rest to the destination with velocity ${{\text{v}}_1}$ and has constant acceleration ${a_1}$. We can write the equation of motion of car A as follows:

${{\text{v}}_1} = {a_1}{t_1}{\text{ }}...{\text{(i)}}$

For car B, let time taken be ${t_2}$, velocity is ${{\text{v}}_2}$ and acceleration is ${a_2}$. Therefore the equation of motion for car B is:

${{\text{v}}_2} = {a_2}{t_2}{\text{ }}...{\text{(ii)}}$

According to the given conditions,

$\begin{gathered} {{\text{v}}_2} = {{\text{v}}_1} - {\text{v }}...{\text{(iii)}} \\\ {t_2} = {t_1} + t{\text{ }}...{\text{(iv)}} \\\ \end{gathered}$

Substituting these values in equation (ii), we get

${{\text{v}}_1} - {\text{v}} = {a_2}\left( {{t_1} + t} \right){\text{ }}...{\text{(v)}}$

Subtracting equation (v) from (i), we get

${\text{v}} = {a_1}{t_1} - {a_2}\left( {{t_1} + t} \right){\text{ }}...{\text{(vi)}}$

Since the total distance travelled by the two cars is equal and cars start from rest (u=0), we can write

$\begin{gathered} {S_A} = {S_B} \\\ \Rightarrow \dfrac{1}{2}{a_1}t_1^2 = \dfrac{1}{2}{a_2}t_2^2 \\\ \end{gathered}$

Using equation (iii) and (iv), we get

$\dfrac{1}{2}{a_1}t_1^2 = \dfrac{1}{2}{a_2}{\left( {{t_1} + t} \right)^2}$

Solving for value of ${t_1}$, we get ${t_1} = \dfrac{{\sqrt {{a_2}} t}}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}$

Substituting this value in equation (vi) and solving for v, we get
${\text{v}} = \left( {\sqrt {{a_1}{a_2}} } \right)t$

Therefore, the correct answer is option D.

Note: All types of translational motion can be described using the three equations of motion. But in these cases, acceleration is always constant. These equations can also be utilized for describing the motion of a body falling under the influence of gravity, where a is substituted by acceleration due to gravity g which has a constant value only when the object is falling near the surface of earth.