Answer:
The value of the constant k is 2
Explanation:
We have the equation of the velocity [tex]v_{x}[/tex] = kt² , where k
is constant and t is the time in second
The particle's position at [tex]t_{0}[/tex] = 0 is [tex]x_{0}[/tex] = -9 m
The particle's position at [tex]t_{1}[/tex] = 3 s is [tex]x_{1}[/tex] = 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 = [tex]\int\ {v_{x} } \, dt[/tex]
Remember in integration we add power by 1 and divide the expression
by the new power
→ x = [tex]\int\ {kt^{2} } \, dt=\frac{1}{3}kt^{3}+c[/tex]
c is the constant of integration to find it substitute the initial value of x
and t in the equation of x
→ [tex]t_{0}[/tex] = 0 , [tex]x_{0}[/tex] = -9 m
→ -9 = [tex]\frac{1}{3}[/tex] k (0)³ + c
→ -9 = c
Substitute the value of c in the equation of x
→ x = [tex]\frac{1}{3}[/tex] k t³ - 9
To find k substitute the values of [tex]t_{1}[/tex] = 3 s , [tex]x_{1}[/tex] = 9 m
→ 9 = [tex]\frac{1}{3}[/tex] k (3)³ - 9
→ 9 = [tex]\frac{1}{3}[/tex] (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
