Question

At time to the position of a body moving
along the s S-axis is s= t^3-6t+9t.
a) Find body's acceleration at each time the
Velocity is zero
b) Find the body's speed at each time the
acceleration is zero
2) Find the total distance traveled by the body
from t=0 to t=2​

Answer 1

Answer:

a..v(0) just gives you the object's initial velocity, which is 3 units/s. To find when the velocity is zero you want to find all t such that 0=t2−4t+3.

b..Find the body's speed each time the acceleration is zero. Since a(t)=6t-12 = 0 when t = 2 and the speed function is sp(t) = |v(t)|, the answer is |v(2)| = 3m/sec.

2..# s = (181sqrt(181)-1000)/27 ~~ 53.15 #

Explanation:

We have parametric equations:

# { (x=5t^2), (y=t^3) :} #

defining the motion of a particle from #t=0# to #t=3#, so the total distance travelled is the arclength, which we calculate for parametric equations using:

# s = int_alpha^beta \ sqrt( (dx/dt)^2+(dy/dt)^2 ) \ dt #

# \ \ = int_0^3 \ sqrt( (10t)^2+(3t^2)^2 ) \ dt #

# \ \ = int_0^3 \ sqrt( t^2(100+9t^2 )) \ dt #

# \ \ = int_0^3 \ tsqrt( 100+9t^2 ) \ dt #

In order to evaluate this integral, we can perform a substitution, Let

# u = 9t^2 +100 => (du)/dt = 18t #

And we change the limits of integration:

# t={ (0), (3) :} => u={ (100), (181) :}#

So then:

# s = int_100^181 \ (1/18) \ sqrt( u ) \ dt #

# \ \ = 1/27 \ (181^(3/2) - 100^(3/2)) #

# \ \ = (181sqrt(181)-1000)/27 #

# \ \ ~~ 53.15 #

Explanation:

thank you ❤️❤️