A particle starts frm a point with a velocity of 6m/s and moves with an acceleration of (-2m/secondsquare). Show that after 6 seconds the particle will beat the starting point ? Plz explain....
s = 6t + (1/2)(-2) t^2 = 6t-t^2
= t(6-t)
s=0 at t=0 and t=6
To show that the particle will return to its starting point after 6 seconds, we need to determine the position function of the particle and evaluate it at t = 0 and t = 6 seconds.
First, let's find the position function of the particle. We know that velocity is the rate of change of position, and acceleration is the rate of change of velocity.
Given that the initial velocity is 6 m/s and the acceleration is -2 m/s², we can integrate the acceleration to obtain the velocity function and then integrate the velocity function to obtain the position function.
Integrating the acceleration function, we get:
v(t) = ∫a(t) dt = ∫(-2) dt = -2t + C
where C is the constant of integration. Since the initial velocity is 6 m/s, we can substitute t = 0 and v = 6 into the equation above to find the value of C:
6 = -2(0) + C
C = 6
Therefore, the velocity function becomes:
v(t) = -2t + 6
Now, we can integrate the velocity function to obtain the position function:
s(t) = ∫v(t) dt = ∫(-2t + 6) dt = -t^2 + 6t + D
where D is the constant of integration. Since the initial position is 0 (starting point), we can substitute t = 0 and s = 0 into the equation above to find the value of D:
0 = -(0)^2 + 6(0) + D
D = 0
Therefore, the position function becomes:
s(t) = -t^2 + 6t
To show that the particle will return to its starting point after 6 seconds, we need to evaluate the position function at t = 6:
s(6) = -(6)^2 + 6(6) = -36 + 36 = 0
The position of the particle at t = 6 seconds is 0, which indicates that the particle returns to its starting point. Thus, after 6 seconds, the particle will beat the starting point.