Question

A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m. Use a numerical method to evaluate the integral.

Answer 1

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation  a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

$\int\limits^a_b {x} \, dx$

a=2

b=1

x=5÷(3s^(1/3)+s^(5/2)) m/s^2

dx=dv

Integrate the left side the standard method.

$\int\limits^a_b {x} \, dx$

a=v

b=0

dx=dv

Integrating

=v^2/2

Use Simpson's rule for the right site.

$\int\limits^a_b {x} \, dx$

a=b

b=a

x=f(x)

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

        =0.8376

Solve for v.  

      v=1.295