Question

A car is moving with speed 30 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.

Answer 1

68 meters moved in the next seconds

Step-by-step explanation:

Given

`u= 30m/s`

`a = 4m/s^2`

Required

Distance covered by the car in the next second

At a point in time t, the current distance is calculated as:

`s(t) = ut + (1)/(2)at^2`

Substitute values for a and u in the above equation.

`s(t) =30 * t + (1)/(2) * 4 * t^2`

`s(t) =30t + 2t^2`

Next, we generate the second degree Taylor polynomial as follows;

Calculate velocity (s'(t))

Differentiate s(t) to get velocity

`s(t) =30t + 2t^2`

`s'(t) =30 + 4t`

Calculate acceleration (s"(t))

Differentiate s'(t) to get acceleration

`s'(t) =30 + 4t`

`s`

When t = 0

We have:

`s(0) = 30 * 0 + 2 * 0^2 = 0`

`s'(0) =30 + 4*0 = 30`

`s`

So, the second degree tailor series is:

`T_2(t) = s(t) * t^0 + s'(t) * (t^1)/(1!) + s`

To see the distance moved in the next second, we set t to 1

So, we have:

`T_2(1) = s(1) * 1^0 + s'(1) * (1^1)/(1!) + s`

`T_2(1) = s(1) * + s'(1) * (1)/(1) + s`

`T_2(1) = s(1) * + s'(1) * 1 + s`

`T_2(1) = s(1) * + s'(1) + (s`

Solving s(1), s'(1) and s"(1)

We have:

`s(1) =30*1 + 2*1^2 = 32`

`s'(1) =30 + 4*1 = 34`

`s`

Hence:

`T_2(1) = 32 + 34 + (4)/(2)`

`T_2(1) = 32 + 34 + 2`

`T_2(1) = 68`