Answer:
Remember that the formula to find the velocity(km/hr) is:
v = \frac{d}{t}v=
t
d
wherein d represents the distance travelled and t is the time interval from when it happened.
Now, the distance covered(d) of the car in the given problem is (2x^3-x^2-4x+3) and the time interval(t) when this happened is (x^2-2x+1) hours.
Using the formula for velocity,
\begin{gathered}v = \frac{d}{t} \\ v = \frac{ {3x}^{3} - {x}^{2} - 4x + 3 }{ {x}^{2} - 2x + 1}\end{gathered}
v=
t
d
v=
x
2
−2x+1
3x
3
−x
2
−4x+3
Using synthetic division for a quicker result,
3 -1 -4 3
-1 | -3 -5
2 | 6 10
----------------------------------
3 5 3 -2
The final answer would be,
v = 3x + 5 \: r. \: 3x - 2v=3x+5r.3x−2
[tex]Hope it helps[/tex]
