Aparticle's velocity is described by the function vx = kt 2 m/s, where k is a constant and t is in s. the particle's position at t0 = 0 s is x0 = -9.0 m. at t1 = 3.0 s, the particle is at x1 = 9.0 m. determine the value of the constant k. be sure to include the proper units.

The value of the constant k is 2

Explanation:

We have the equation of the velocity = kt² , where k

is constant and t is the time in second

The particle's position at = 0 is = -9 m

The particle's position at = 3 s is = 9 m

We need to find the value of the constant k

The relation between the velocity and the displacement in a particular

time is x =

Remember in integration we add power by 1 and divide the expression

by the new power

→ x =

c is the constant of integration to find it substitute the initial value of x

and t in the equation of x

→ = 0 , = -9 m

→ -9 = k (0)³ + c

→ -9 = c

Substitute the value of c in the equation of x

→ x = k t³ - 9

To find k substitute the values of   = 3 s , = 9 m

→ 9 = k (3)³ - 9

→ 9 = (27) k - 9

→ 9 = 9 k - 9

Add 9 to both sides

→ 18 = 9 k

Divide both sides by 9

→ k = 2

The value of the constant k is 2

cancer risk hope this

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..