Given:
Following data is given about a moving particle
a(t) = 2t + 9, s(0) = 6, v(0) = −9
Find:
we have to find the position s(t) of the particle.
Explanation:
Acceleration of the particle is given as
a(t) = 2t + 9,
Integrate a(t) wtih respect to t, we get
∫a(t)dt = ∫(2t + 9)dt
v(t) = t² + 9t + c, where c is a constant
given v(0) = -9
Therefore, we get
v(0) = (0)² + 9(0) + c
-9 = c
c = -9
Therefore, v(t) = t² + 9t - 9
Now integrate v(t), we get
∫v(t)dt = ∫(t² + 9t - 9)dt
s(t) = (t³)/3 + (9t²)/2 - 9t + d, where d is a constant
given s(0) = 6
This implies, s(0) = (0³)/3 + (9(0)²)/2 - 9(0) + d
6 = d
Therefore, the position of the particle at any time t is
[tex]s(t)=\frac{t^3}{3}+\frac{9t^2}{2}-9t+6[/tex]
