Question #286367

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.


1
Expert's answer
2022-01-11T12:49:04-0500

a)


V=5t2+2t4t2V=5t^2+2t-\dfrac{4}{t^2}

distance=25(5t2+2t4t2)dtdistance=\displaystyle\int_{2}^{5}(5t^2+2t-\dfrac{4}{t^2})dt

=[5t33+t2+4t]52=\big[\dfrac{5t^3}{3}+t^2+\dfrac{4}{t}\big]\begin{matrix} 5 \\ 2 \end{matrix}

=5(5)33+(5)2+45(5(2)33+(2)2+42)=\dfrac{5(5)^3}{3}+(5)^2+\dfrac{4}{5}-(\dfrac{5(2)^3}{3}+(2)^2+\dfrac{4}{2})

=214.8(units)=214.8 (units)

b) The Simpson's 1/3 rule

Approximate the integral 25(5t2+2t4t2)dt\displaystyle\int_{2}^{5}(5t^2+2t-\dfrac{4}{t^2})dt with n=6n=6  using the Simpson's rule.


abV(t)dt=Δt3(V(t0)+4V(t1)+2V(t2)+...\displaystyle\int_{a}^{b}V(t)dt=\dfrac{\Delta t}{3}(V(t_0)+4V(t_1)+2V(t_2)+...

+2V(tn2)+4V(tn1)+V(tn))+2V(t_{n-2})+4V(t_{n-1})+V(t_n))

We have that  V(t)=5t2+2t4t2,a=2,b=5,n=6V(t)=5t^2+2t-\dfrac{4}{t^2}, a=2, b=5, n=6

Then Δt=ban=526=0.5\Delta t=\dfrac{b-a}{n}=\dfrac{5-2}{6}=0.5

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

Now, just evaluate the function at these endpoints.


V(t0)=V(2)=23V(t_0)=V(2)=23

4V(t1)=4V(2.5)=142.444V(t_1)=4V(2.5)=142.44

2V(t2)=2V(3)=91092V(t_2)=2V(3)=\dfrac{910}{9}

4V(t3)=4V(3.5)=13313494V(t_3)=4V(3.5)=\dfrac{13313}{49}

2V(t4)=2V(4)=175.52V(t_4)=2V(4)=175.5

4V(t5)=4V(4.5)=35657814V(t_5)=4V(4.5)=\dfrac{35657}{81}

V(t6)=V(5)=134.84V(t_6)=V(5)=134.84

16(23+142.44+9109+1331349+175.5\dfrac{1}{6}(23+142.44+\dfrac{910}{9}+\dfrac{13313}{49}+175.5

+3565781+134.84)214.7991(units)+\dfrac{35657}{81}+134.84)\approx 214.7991(units)

Theanswer for distance from part A 214.8214.8 units and from part B 214.799214.799 units are very close


214.799214.8214.8×100%=0.0005%\dfrac{|214.799-214.8|}{214.8}\times 100\%=0.0005\%


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