Answer:
68 meters moved in the next seconds
Explanation:
Given
[tex]u= 30m/s[/tex]
[tex]a = 4m/s^2[/tex]
Required
Distance covered by the car in the next second
At a point in time t, the current distance is calculated as:
[tex]s(t) = ut + \frac{1}{2}at^2[/tex]
Substitute values for a and u in the above equation.
[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]
[tex]s(t) =30t + 2t^2[/tex]
Next, we generate the second degree Taylor polynomial as follows;
Calculate velocity (s'(t))
Differentiate s(t) to get velocity
[tex]s(t) =30t + 2t^2[/tex]
[tex]s'(t) =30 + 4t[/tex]
Calculate acceleration (s"(t))
Differentiate s'(t) to get acceleration
[tex]s'(t) =30 + 4t[/tex]
[tex]s"(t) =4[/tex]
When t = 0
We have:
[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]
[tex]s'(0) =30 + 4*0 = 30[/tex]
[tex]s"(0) = 4[/tex]
So, the second degree tailor series is:
[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]
To see the distance moved in the next second, we set t to 1
So, we have:
[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]
Solving s(1), s'(1) and s"(1)
We have:
[tex]s(1) =30*1 + 2*1^2 = 32[/tex]
[tex]s'(1) =30 + 4*1 = 34[/tex]
[tex]s"(1) =4[/tex]
Hence:
[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]
[tex]T_2(1) = 32 + 34 + 2[/tex]
[tex]T_2(1) = 68[/tex]
