Question

KHW 1 Item 16 Part A The as graph for a rocket moving along a straight track has been experimentally determined The rocket starts at 8-0 when u= 0 Figure 1) Determine its speed when it is at s 85 ft Express your answer to three significant figures and include the appropriate units 292 ft s An Correct Part B Determine its speed when it is at s 135 ft. Use Simpson's rule with 100 Express your answer to three significant figures and include the appropriate units. Figure 1 of 1 a (ft/s) ft 20.05 Submit a- 5+6s-10)3 X Incorrect Try Again; 3 attempts remaining s (ft) 100 Provide Feedback

Answer 1

b) here we have

0<s<100

from kinematics law

vdv=ads

$\int_{0}^{v}$vdv=$\int_{0}^{s}$5ds

v²/2=5s

v²=10s

v=$\sqrt{}$10s

since the maximum limit is 100

at s=100

v=$\sqrt{}$10*100=31.623

so now again using

vdv=ads

31.623
vdv=
r135 100
[5+6($\sqrt{}$s-10)^5/3]ds....................a)  

lets do the right side of the integral because its more complex

r135 100
[5+6($\sqrt{}$s- 10)^1.6667ds=6$\int$($\sqrt{}$s-10)^1.6667+5$\int$1ds

let u=$\sqrt{}$x-10,

dx=2$\sqrt{}$xdu

=2$\int$u^2.6667+10u^1.6667du..................1)

=$\int$u^2.6667+10$\int$u^1.6667du

=10000u^36667/1000/3667+10*10000u^26667/10000/26667

we have 2 multiply it by 2 since in eqn 1 we have 2 outside

=20000u^36667/10000/36667+200000u^26667/10000/26667

now putting u=$\sqrt{}$x-10 from earlier

=20000($\sqrt{}$x-10)^36667/10000/36667+200000($\sqrt{}$x-10)^26667/10000/26667

now solving integration of 1ds, it will be just s, so now the whole term will become

=$\int$6($\sqrt{}$x-10)^1.6667 ds+5$\int$1ds

=5s+120000($\sqrt{}$x-10)^3667/1000/3667+400000($\sqrt{}$x-10)^2667/10000

on solving this we get=356.7655

now back to the whole intergral from eqn a)

v²/2|^v_31.623=356.7655

1578031v²-1578053296/3156062=356.7655

1578031v²=(356.772*3156062)+1578053296

v²=1713.5581

v=41.4 ft/s

so the answer is 41.4 ft/s