Physics, asked by ayushgupta6599, 10 months ago

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'?

Answers

Answered by ShivamKashyap08
15

\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:-}}}

\rule{300}{1.5}

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}}}

\rule{300}{1.5}

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}}}

\rule{300}{1.5}

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}}}}

\rule{300}{1.5}

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}}}}

\rule{300}{1.5}

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)}

\rule{300}{1.5}

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)}

\rule{300}{1.5}

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".
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