The position of a function of a moving particle is s(t)=5+4t-t^2 for 0<t<10 where s is in meters and t is measured in seconds. What is the maximum speed in m/sec of the particle on the interval [0,10]?
v(t) = 4-2t
dv/dt = -2
There is no local min/max, and v is constantly decreasing, so max v in the interval is at t=0, or v=4
To find the maximum speed of the particle on the interval [0,10], we need to find the derivative of the position function with respect to time.
The position function is given as:
s(t) = 5 + 4t - t^2
To find the derivative, we differentiate the function with respect to time (t):
s'(t) = 4 - 2t
Now, to find the maximum speed, we need to find the maximum value of the absolute value of the derivative within the interval [0,10].
Let's first find the critical points by setting the derivative equal to zero:
4 - 2t = 0
Solving for t:
2t = 4
t = 2
Now, let's evaluate the absolute value of the derivative at the critical points and the endpoints of the interval:
s'(0) = 4 - 2(0) = 4
s'(2) = 4 - 2(2) = 0
s'(10) = 4 - 2(10) = -16
The maximum speed is determined by the maximum absolute value of the derivative.
Since the absolute value of s'(10) is greater than the absolute values of s'(0) and s'(2), the maximum speed is |s'(10)| = |-16| = 16 m/sec.
Therefore, the maximum speed of the particle on the interval [0,10] is 16 m/sec.
To find the maximum speed of the particle, we need to find the derivative of the position function, which gives us the velocity function. Then, we can determine the maximum value of the velocity function over the given interval.
First, let's find the derivative of the position function, s(t).
s(t) = 5 + 4t - t^2
To find the derivative, we differentiate each term with respect to time (t):
s'(t) = 0 + 4 - 2t
Simplifying further, we get:
s'(t) = 4 - 2t
Now, the velocity function, v(t), represents the rate of change of position with respect to time. In this case, it is the derivative of the position function:
v(t) = s'(t) = 4 - 2t
To determine the maximum speed, we need to find the maximum value of the velocity function within the given interval [0,10].
Since the velocity function is a linear function with a negative slope (-2t), the maximum value occurs at the start of the interval (t=0) or the end of the interval (t=10).
Let's calculate the velocity at both endpoints:
v(0) = 4 - 2(0) = 4
v(10) = 4 - 2(10) = 4 - 20 = -16
The maximum speed will be the greater of these two values, so the maximum speed is 4 m/s.
Therefore, the maximum speed of the particle on the interval [0,10] is 4 m/s.