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 when x = 8m using kinematic equation 3, you need to find the integral of acceleration (a) with respect to displacement (x). However, to properly evaluate the integral, you need to set up the limits of integration.
In this case, you want to find the velocity when x = 8m. You can set up the integral from the initial position (x = 1m, which is given) to the final position (x = 8m). So the integral becomes:
∫(a dx) = ∫[(4x + 5x^(3/2))] dx
To solve this integral, you can break it down into two separate integrals:
∫(4x dx) + ∫(5x^(3/2) dx)
Integrating the first term with respect to x:
∫(4x dx) = 2x^2 + C₁
And integrating the second term with respect to x:
∫(5x^(3/2) dx) = (10/5)(2/5)x^(5/2) + C₂
= (4/5)x^(5/2) + C₂
Combining the two integrals:
∫[(4x + 5x^(3/2))] dx = 2x^2 + (4/5)x^(5/2) + C₁ + C₂
= 2x^2 + (4/5)x^(5/2) + C
Where C is the constant of integration.
To evaluate the velocity when x = 8m, substitute this value into the integral expression and solve for the velocity (v):
v = 2(8^2) + (4/5)(8^(5/2)) + C
Now, without more information about the constant of integration (C), we cannot determine the exact velocity value. However, you can calculate an approximate value by substituting a reasonable value for C into the equation.