Skateboarders Velocity is given as
v=3t-0.2t^2
What is his max velocity
To get the max velocity, first, differentiate the velocity equation with respect to time. Then equate to zero:
dv/dt = 3 - 0.4t
0 = 3 - 0.4t
0.4t = 3
t = 7.5
Substitute to the original equation:
v = 3t-0.2t^2
v = 3(7.5) - 0.2(7.5)^2
v = ?
Or you may also want to plot the equation. You should get a parabola which has a maximum. That max point corresponds to the velocity (y-coordinate) at a particular time (x-coordinate).
Hope this helps~ :)
THANKS MATE !
To find the maximum velocity, we need to determine the critical points of the velocity function. In this case, the velocity function is given by v = 3t - 0.2t^2.
To find the critical points, we need to find the values of t where the derivative of the velocity function equals zero. Differentiating the velocity function with respect to t, we get:
dv/dt = 3 - 0.4t
To find the critical points, we set the derivative equal to zero and solve for t:
3 - 0.4t = 0
0.4t = 3
t = 3/0.4
t = 7.5
We have found that the critical point is t = 7.5. To determine whether it is a maximum or minimum, we can take the second derivative:
d^2v/dt^2 = -0.4
Since the second derivative is negative, it means that the critical point represents a maximum. Therefore, the maximum velocity occurs when t = 7.5.
To find the corresponding velocity, substitute t = 7.5 back into the velocity function:
v = 3(7.5) - 0.2(7.5)^2
v = 22.5 - 0.2(56.25)
v = 22.5 - 11.25
v = 11.25
Hence, the maximum velocity of the skateboarder is 11.25 units per time.