Question

A particle'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.

Answer 1

Answer:

The value of the constant k is 2

Explanation:

We have the equation of the velocity $v_{x}$ = kt² , where k

is constant and t is the time in second

The particle's position at $t_{0}$ = 0 is $x_{0}$ = -9 m

The particle's position at $t_{1}$ = 3 s is $x_{1}$ = 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 = $\int\ {v_{x} } \, dt$

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

by the new power

→ x = $\int\ {kt^{2} } \, dt=\frac{1}{3}kt^{3}+c$

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

and t in the equation of x

→ $t_{0}$ = 0 , $x_{0}$ = -9 m

→ -9 = $\frac{1}{3}$ k (0)³ + c

→ -9 = c

Substitute the value of c in the equation of x

→ x = $\frac{1}{3}$ k t³ - 9

To find k substitute the values of  $t_{1}$ = 3 s , $x_{1}$ = 9 m

→ 9 = $\frac{1}{3}$ k (3)³ - 9

→ 9 = $\frac{1}{3}$ (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