A car is moving with speed 60 m/s and acceleration 4 m/s2 at a given instant. (a) using a second-degree taylor polynomial, estimate how far the car moves in the next second. m (b) would it be reasonable to use this polynomial to estimate the distance traveled during the next minute? yes no

b) The second order Taylor polynomial will be reasonable to estimate the distance traveled during the next minute if and only if the initial acceleration remains constant in time.

a) Let the position of the car be represented by a function in terms of time:

[tex]x = f(t)[/tex] (1)

The position of the car in time can be estimated by means of a second-degree Taylor polynomial:

[tex]x = x_{o} + \frac{v_{o}}{1!} \cdot t + \frac{a_{o}}{2!}\cdot t^{2}[/tex]

[tex]x = x_{o}+v_{o}\cdot t + \frac{1}{2}\cdot a_{o}\cdot t^{2}[/tex] (2)

Where:

[tex]x_{o}[/tex] - Initial position, in meters.
[tex]v_{o}[/tex] - Initial speed, in meters per second.
[tex]a_{o}[/tex] - Initial acceleration, in meters per square second.
[tex]t[/tex] - Time, in seconds.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 60\,\frac{m}{s}[/tex], [tex]a_{o} = 4\,\frac{m}{s^{2}}[/tex] and [tex]t = 1\,s[/tex], then the distance traveled in the next second is:

[tex]x = 0\,m + \left(60\,\frac{m}{s} \right)\cdot (1\,s) +\frac{1}{2}\cdot \left(4\,\frac{m}{s^{2}} \right) \cdot (1\,s)^{2}[/tex]

[tex]x = 62\,m[/tex]

The estimated distance travelled by the car in the next second is 62 meters.

b) The second order Taylor polynomial will be reasonable to estimate the distance traveled during the next minute if and only if the initial acceleration remains constant in time.

We kindly invite to check this question on Taylor polynomials: https://brainly.com/question/15581839