Question

The acceleration of a moving object can be modelled by the following

equation.

a = 3t^2

i) Using integration v = ∫ a dt to find the value of the velocity value of

the velocity at 0 ≤ t ≤ 2.2 (s).

ii) Using trapezium rules (n=6) to find the mean value of the velocity over

a range of 0 ≤ t ≤ 2.2.

iii) Using a computer spreadsheet to increase the number of intervals to

n=12 for the numerical method and compare your obtained results

with (i) and (ii).

iv) Evaluate all obtained results (form i to iv steps) using technically correct

language and a logical structure to identify whether the size of

numerical steps (intervals) and applied various numerical techniques

have affected the obtained result.

Answer 1

$a = 3t²$

(i) $v = \int a\ dt = \int3t²\ dt = t³$

$v(0) = 0³ = 0m/s\\ v(2.2) = 2.2³ = 10.648m/s$

(ii) $\int^{2.2}_{{{0}}}3t² \ dt$

$f(x) = 3t², \quad a={ {{0}} } , \quad b = 2.2$

The width of each subinterval is

$Δ x = \dfrac{b − a}{ n} = \dfrac{2.2}{ 6}=0.37$

so the grid points have the coordinates

$x _ i = i 0.37$

Calculate the values of the function f(x) at the points x_i:

${f\left( {{x_0}} \right) }={ f\left( 0 \right) }={ {3}{( {{0 ²)}} }}={ {0^2} }={ 0}$

${f\left( {{x_1}} \right) }={ f\left( {0.37} \right) }={ {3}(0.37²}) = 0.4107$

${f\left( {{x_2}} \right) }={ f\left( {2(0.37)} \right) }={ {3}(0.74²}) = 1.6428$

${f\left( {{x_3}} \right) }={ f\left( {3(0.37)} \right) }={ {3}(1.11²}) = 3.6963$

${f\left( {{x_4}} \right) }={ f\left( {4(0.37)} \right) }={ {3}(1.48²}) = 6.5712$

${f\left( {{x_5}} \right) }={ f\left( {6(0.37)} \right) }={ {3}(1.85²}) = 10.2675$

${f\left( {{x_6}} \right) }={ f\left( {2(0.37)} \right) }={ {3}(2.22²}) = 14.7852$

The Trapezoidal Rule formula is written in the form;

${\int^{2.2}_{{{0}}}3t²\ dt \approx {T_6} }={ \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots }\right.}+{\left.{ 2f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right] }={ \frac{0.37}{{2}}\left[ {0 + 2 \cdot 0.4107 + 2 \cdot 1.6428 }\right.}+{\left.{ 2 \cdot 3.6963 + 2 \cdot 6.5712 + 2 \cdot 10.2675 + 2\cdot14.7852} \right] }=$

$19.2986\ m/s$

(iii) $\int^{2.2}_{{ {0} }} a\ dt = \int^{2.2}_{{ {0} }}3t²\ dt = [t³]^{2.2}_{ {{0} }} = 2.2³ - 0³ = 10.648 \ m/s$

which coincides with the original value

(iv) all numerical steps have affected the original obtained result.