(a) If the position of a particle is given by x = 16t - 7t3, where t is in seconds and x is in meters, when is the particle's velocity zero? (b) When is its acceleration a zero?
Please visit website yougems e-learning (educational search engine) a reliable and trustworthy site to get answers free by experts in defined period. Thanks
To find when the particle's velocity is zero, we need to determine the value(s) of time (t) when the derivative of the position function, x(t), with respect to time is zero. Similarly, to find when the acceleration is zero, we need to determine the time(s) when the derivative of the velocity function, v(t), is zero.
Let's start by finding the velocity function by taking the derivative of the position function with respect to time:
x(t) = 16t - 7t^3
To find the velocity function, we differentiate x(t) with respect to t:
v(t) = d/dt(16t - 7t^3)
Now let's apply the power rule of differentiation:
v(t) = 16 - 21t^2
(a) To find when the velocity is zero, we set v(t) equal to zero and solve for t:
0 = 16 - 21t^2
Rearranging the equation, we have:
21t^2 = 16
Next, divide both sides of the equation by 21:
t^2 = 16/21
Taking the square root of both sides, we get:
t = ±√(16/21)
Thus, the particle's velocity is zero when t is approximately ±0.802 seconds.
(b) To find when the acceleration is zero, we need to differentiate the velocity function, v(t), with respect to time:
a(t) = d/dt(16 - 21t^2)
Differentiating, we have:
a(t) = -42t
Now, we set a(t) equal to zero and solve for t:
0 = -42t
Dividing both sides by -42:
t = 0
Thus, the particle's acceleration is zero at t = 0 seconds.