Question

The velocity of a moving object is given by the function

𝒅𝑽=𝟓𝒕^2+2t-4/t^2

a) Integrate the function and find the distance travelled in metres between t=2 and t=5

b) Calculate the distance travelled using a numerical method and compare your answer with part A.

Answer 1

a)

$$V=5t^2+2t-\dfrac{4}{t^2}$$

$$distance=\displaystyle\int_{2}^{5}(5t^2+2t-\dfrac{4}{t^2})dt$$

$$=\big[\dfrac{5t^3}{3}+t^2+\dfrac{4}{t}\big]\begin{matrix} 5 \\ 2 \end{matrix}$$

$$=\dfrac{5(5)^3}{3}+(5)^2+\dfrac{4}{5}-(\dfrac{5(2)^3}{3}+(2)^2+\dfrac{4}{2})$$

$$=214.8 (units)$$

b) The Simpson's 1/3 rule

Approximate the integral $\displaystyle\int_{2}^{5}(5t^2+2t-\dfrac{4}{t^2})dt$ with $n=6$  using the Simpson's rule.

$$\displaystyle\int_{a}^{b}V(t)dt=\dfrac{\Delta t}{3}(V(t_0)+4V(t_1)+2V(t_2)+...$$

$$+2V(t_{n-2})+4V(t_{n-1})+V(t_n))$$

We have that  $V(t)=5t^2+2t-\dfrac{4}{t^2}, a=2, b=5, n=6$

Then $\Delta t=\dfrac{b-a}{n}=\dfrac{5-2}{6}=0.5$

Divide the interval $[2,5]$ into $n=6$ subintervals of the length $\Delta t=0.5$ with the following endpoints: $a=2, 2.5, 3, 3.5, 4, 4.5, 5=b.$

Now, just evaluate the function at these endpoints.

$$V(t_0)=V(2)=23$$

$$4V(t_1)=4V(2.5)=142.44$$

$$2V(t_2)=2V(3)=\dfrac{910}{9}$$

$$4V(t_3)=4V(3.5)=\dfrac{13313}{49}$$

$$2V(t_4)=2V(4)=175.5$$

$$4V(t_5)=4V(4.5)=\dfrac{35657}{81}$$

$$V(t_6)=V(5)=134.84$$

$$\dfrac{1}{6}(23+142.44+\dfrac{910}{9}+\dfrac{13313}{49}+175.5$$

$$+\dfrac{35657}{81}+134.84)\approx 214.7991(units)$$

Theanswer for distance from part A $214.8$ units and from part B $214.799$ units are very close

$$\dfrac{|214.799-214.8|}{214.8}\times 100\%=0.0005\%$$