The acceleration of a particle which moves along the x axis is given by a=4x+5x^(3/2)
The particle has speed of 2m/s when x=1m
x is in meters
a is in m/s^2
evaluate the velocity when x = 8m
I know that I need to use kinematic eqn 3
adr = vdv
then take the integral but I'm not sure where to take the limits from
To evaluate the velocity of the particle when x = 8m, you can use the kinematic equation 3, which is:
adt = vdv
To solve this equation, you need to integrate both sides. However, before integrating, let's express the given acceleration equation in terms of velocity.
Given acceleration, a = 4x + 5x^(3/2)
Since we know that v = dx/dt (velocity is the derivative of displacement with respect to time), we can rewrite the acceleration equation in terms of velocity by substituting x with v:
a = 4v + 5v^(3/2)
Now, let's integrate both sides of the modified acceleration equation:
∫ a dt = ∫ (4v + 5v^(3/2)) dt
Integrating the left side with respect to time (t) gives us:
∫ a dt = ∫ 0 dt = C1
Integrating the right side with respect to velocity (v) gives us:
∫ (4v + 5v^(3/2)) dt = ∫ (4v + 5v^(3/2)) dv
∫ (4v + 5v^(3/2)) dv = (2v^2 + (10/5)(2/5)v^(5/2)) + C2
where C1 and C2 are constants of integration.
Now, we know that when the speed is 2 m/s (v = 2 m/s), and x = 1 m, we can substitute these values into the equation to find the constants of integration (C1 and C2).
For the first condition, when x = 1 m and v = 2 m/s:
2 = 2(1)^2 + (10/5)(2/5)(1)^(5/2) + C2
2 = 2 + 2/5 + C2
2 - 2 - 2/5 = C2
C2 = -2/5
Therefore, C2 is -2/5.
Now, we substitute C2 into the integrated equation:
∫ (4v + 5v^(3/2)) dv = (2v^2 + (10/5)(2/5)v^(5/2)) - 2/5
So, the equation becomes:
∫ a dt = (2v^2 + (10/5)(2/5)v^(5/2)) - 2/5 = C1
Now, we have the integrated form of the equation, and we can use this to evaluate the velocity when x = 8 m.
To do this, we substitute x = 8 m into the equation:
(2v^2 + (10/5)(2/5)v^(5/2)) - 2/5 = C1
Now we need to solve for v.