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Step 1 : First find the distance of the car traveled by differentiating it

$\frac{d}{dx}$ ( x^{2}-5x+6)

y1= 2x - 5

y1= 2(5) - 5

= 10 - 5

= 5 m**2.** Find the acceleration of the body y= x^{3} - 2x at x = 1m

*Step 1*

First find the derivative for the above equation

Y1 = 3x^{2} - 2

Y2 = 6x

= 6(1)

=6 m/sec2.

Y= = 2x

$\frac{dy}{dx}$ = 4x + 4

= 4(-2) + 4

= -8 + 4

= -4 at (x y)

Now the equation of tangent is given by y - y

y - (-3) = -4(x - (-2))

y + 3 = -4x - 8

y + 3 + 4x + 8 = 0

y + 4x + 11 = 0

Now to find the equation of normal is given by y-y

y-(-3) = -$\frac{1}{4(x-(-2))}$

4y + 3 = -x - 2

4y + x + 5 = 0 is the equation of the normal