A particle moves along the x-axis so that at any time t≥0, its velocity is given by v(t)=t^2-16t+4 What is the velocity of the particle when its acceleration is zero?
since a(t) = 2t - 16
when a(t) = 0
t = 8
v(8) = 64 - 16(8) + 4 = ......
To find the velocity of the particle when its acceleration is zero, we need to determine the value of t when the acceleration is zero and then substitute that value into the expression for velocity.
Acceleration is defined as the derivative of velocity with respect to time. So, let's find the acceleration by taking the derivative of the given velocity function v(t):
a(t) = v'(t) = (t^2 - 16t + 4)'
To find the derivative, we can apply the power rule and the constant rule of differentiation:
a(t) = 2t - 16
Now, we need to find the value of t when the acceleration is zero. Set the acceleration function equal to zero and solve for t:
2t - 16 = 0
2t = 16
t = 8
So, when t = 8, the particle's acceleration is zero.
Finally, let's substitute this value of t into the original velocity function to find the velocity:
v(t) = t^2 - 16t + 4
v(8) = 8^2 - 16(8) + 4
v(8) = 64 - 128 + 4
v(8) = -60
Therefore, the velocity of the particle when its acceleration is zero is -60.