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)

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.

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