Question

A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate how far the car moves in the next second. Do you think it would be reasonable to use this polynomial to estimate the distance traveled during the next minute? Why or why not?

Answer 1

Answer:

$T(1)=21$

Explanation:

The equation of the position in kinematics is given:

$x(t)=x_{0}+v_{0}t+0.5at^{2}$

• x(0) is the initial position, in this it is 0
• v(0) is the initial velocity (20 m/s)
• a is the acceleration (2 m/s²)

So the equation will be:

$x(t)=20t+0.5*2*t^{2}$

$x(t)=20t+t^{2}$    

Now, the Taylor polynomial equation is:

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

$f(t)=x(t)=20t+t^{2}$

$f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t$

$f''(t)=\frac{dv(t)}{dt}=a(t)=2$

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

$f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}$

$f(0)=x(0)=0$

$f'(0)=v(0)=20$

$f''(0)=a(0)=2$

$T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}$

$T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}$

$T(t)=20t+t^{2}$

Let's put t=1 so find the how far the car moves in the next second:

$T(1)=20*1+1^{2}$

$T(1)=21$

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!