Question

1. what is the speed of a runner whose distance is represented by the function ( d ( h ) = 6 h ^ { 2 } - h - 1 ) and whose time is represented by ( t ( h ) = 2 h - 1 )?

( ( \frac { d } { t } ) ( h ) = 12 h ^ { 3 } - 8 h ^ { 2 } - h + 1 )
( ( \frac { d } { t } ) ( h ) = 3 h + 1 )
( ( \frac { d } { t } ) ( h ) = 3 h + 1 + \frac { 3 } { 2 h - 1 } )
( ( \frac { d } { t } ) ( h ) = 12 h ^ { 3 } + 2 h ^ { 2 } - 2 h )

Explanation:

Step1: Recall the formula for speed

Speed \(v=\frac{d}{t}\), where \(d\) is distance and \(t\) is time. Here, \(d(h) = 6h^{2}-h - 1\) and \(t(h)=2h - 1\).

Step2: Perform polynomial long - division

Divide \(6h^{2}-h - 1\) by \(2h - 1\).
First, divide the leading term of the dividend \(6h^{2}\) by the leading term of the divisor \(2h\): \(\frac{6h^{2}}{2h}=3h\).
Multiply \(3h\) by \((2h - 1)\): \(3h\times(2h - 1)=6h^{2}-3h\).
Subtract this from the dividend: \((6h^{2}-h - 1)-(6h^{2}-3h)=2h - 1\).
Then divide the new leading term \(2h\) by \(2h\) (from the divisor): \(\frac{2h}{2h}=1\).
Multiply \(1\) by \((2h - 1)\): \(2h - 1\).
Subtract: \((2h - 1)-(2h - 1)=0\).

Answer:

\((\frac{d}{t})(h)=3h + 1\)