Question

In a car race on straight road, car a takes a time t less than car b at the finish and passes finishing point with a speed 'v' more than that of car
b. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then find the magnitude of v'?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Acceleration of 1st car = a1.
• Acceleration of 2nd car = a2.
• Car "A" is faster than Car "B" by "v".
• Let the distance travelled by both the car be "s".

$\huge{\bold{\underline{Explanation:-}}}$

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For Car "A".

As initial velocity (u) = 0 m/s.

And the velocity of the car be ${v_a}$

Applying third kinematic equation,

$\large{\bold{ \tt v^2 - u^2 = 2as}}$

Substituting the values,

$\large{(v_a)^2 - 0 = 2a_1s}$

$\large{v_a = \sqrt{2a_1s}}$

$\large{\boxed{ \tt v_a = \sqrt{2a_1s}}}$

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For Car "B".

As initial velocity (u) = 0 m/s.

And the velocity of the car be ${v_n}$

Applying third kinematic equation,

$\large{\bold{ \tt v^2 - u^2 = 2as}}$

Substituting the values,

$\large{(v_b)^2 - 0 = 2a_2s}$

$\large{v_b = \sqrt{2a_2s}}$

$\large{\boxed{ \tt v_b = \sqrt{2a_2s}}}$

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For Car "A"

Time taken = ?.

Applying second kinematic equation,

$\large{\bold{ \tt S = ut + \dfrac{1}{2}at^2}}$

As initial velocity is zero.

$\large{S = \dfrac{1}{2}at^2}$

Let the time taken be ${t_a}$

Now,

$\large{s = \dfrac{1}{2}(a_1)(t_a)^2}$

Now,

$\large{\boxed{ \tt t_a = \sqrt{ \dfrac{2s}{a_1}}}}$

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For Car "B"

Time taken = ?.

Applying second kinematic equation,

$\large{\bold{ \tt S = ut + \dfrac{1}{2}at^2}}$

As initial velocity is zero.

$\large{S = \dfrac{1}{2}at^2}$

Let the time taken be ${t_b}$

Now,

$\large{s = \dfrac{1}{2}(a_2)(t_b)^2}$

Now,

$\large{\boxed{ \tt t_b = \sqrt{ \dfrac{2s}{a_2}}}}$

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As the Velocity of "A" is greater than velocity of "B" "v".

Therefore,

$\large{\bold{ \tt v = v_a - v_b}}$

Substituting the values,

$\large{v = \sqrt{2a_1s} - \sqrt{2a_2s}}$

Now,

$\large{\boxed{ \tt v = \sqrt{2s}( \sqrt{a_1} - \sqrt{a_2})}}$

$\large{ \tt v = \sqrt{2s}( \sqrt{a_1} - \sqrt{a_2}) \: ----(1)}$

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As velocity of "A" is greater than the velocity of "B" .

Then Car "A" will take less time.

Now,

$\large{\bold{ \tt t = t_a - t_b}}$

$\large{t = \sqrt{ \dfrac{2s}{a_1}} - \sqrt{ \dfrac{2s}{a_2}}}$

Now,

$\large{t = \sqrt{2s} \left[ \dfrac{1}{ \sqrt{a_1}} - \dfrac{1}{ \sqrt{a_2}} \right]}$

Taking L.C.M.

$\large{\boxed{ \tt t = \sqrt{2s} \left[ \dfrac{ \sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1a_2}} \right]}}$

$\large{ \tt t = \sqrt{2s} \left[ \dfrac{ \sqrt{a_1} - \sqrt{a_2}}{\sqrt{a_1a_2}} \right]\: ----(2)}$

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Now, doing,

$\large{ \dfrac{Equation (1)}{Equation (2)}}$

$\large{ \dfrac{v}{t} = \dfrac{ \sqrt{2s} (\sqrt{a_1} - \sqrt{a_2})}{ \dfrac{ \sqrt{2s} (\sqrt{a_1} - \sqrt{a_2}}{ \sqrt{a_1a_2}}}}$

$\large{ \dfrac{v}{t} = \dfrac{ \cancel{\sqrt{2s}} \cancel{(\sqrt{a_1} - \sqrt{a_2})}}{ \dfrac{ \cancel{\sqrt{2s}} \cancel{(\sqrt{a_1} - \sqrt{a_2})}}{ \sqrt{a_1a_2}}}}$

It comes as,

$\large{ \dfrac{v}{t} = \sqrt{a_1a_2}}$

Now,

$\huge{\boxed{\boxed{ \tt v = t. \sqrt{a_1}{a_2}}}}$

Note:-

• Here "t" is the time difference between the Car"A" and Car"B".