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Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_p \left( t \right) = at + bt^2$

and $x_Q \left( t \right) = ft - t^2$ . At what time do the cars have the same velocity?

Option 1)
$\frac{{a - f}}{{1 + b}}$

Option 2)
$\frac{{a + f}}{{2\left( {b - 1} \right)}}$

Option 3)
$\frac{{a + f}}{{2\left( {1 + b} \right)}}$

Option 4)
$\frac{{f - a}}{{2\left( {1 + b} \right)}}$

Answers (1)

best_answer

As we learned in

Introduction to Differentiation -

$frac{d}{dx}left ( x^{n} ight )={n} x^{n-1}$

- wherein

$frac{d}{dx}left ( x^{5} ight )=left ( n=5 ight )$

$Rightarrow {n}x^{n-1}$

$Rightarrow {5}x^{5-1}$

$Rightarrow {5}x^{4}$

$X_{P}(t)=at+bt^{2},\:X_{Q}(t)=ft-t^{2}$

$V_{P}=a+2bt$ $V_{Q}= f-2t$

$\Delta r$ $V_{P}=V_{Q}$

$a+2bt=f-2t$

$t = \frac{f-a}{2(b+1)}$

Option 1)

$\frac{{a - f}}{{1 + b}}$

This option is incorrect.

Option 2)

$\frac{{a + f}}{{2\left( {b - 1} \right)}}$

This option is incorrect.

Option 3)

$\frac{{a + f}}{{2\left( {1 + b} \right)}}$

This option is incorrect.

Option 4)

$\frac{{f - a}}{{2\left( {1 + b} \right)}}$

This option is correct.

Posted by

divya.saini

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