A model rocket is launched straight upward. Its altitude y as a function of time is given
by y = btct2
, where b = 82 m/s, c = 4.9 m/s2
is the time in seconds, and y is in meters. (a) Use
differentiation to find a general expression for the rockets velocity as a function of time. (b) When
is the velocity zero?
To find the velocity of the rocket as a function of time, we can differentiate the equation for the altitude with respect to time.
Given: y = bt - ct^2
(a) Taking the derivative of y with respect to time (t):
dy/dt = d(bt - ct^2)/dt
To differentiate the equation, we can use the power rule and the constant rule of differentiation.
The power rule states that if we differentiate x^n with respect to x, the result is nx^(n-1). And the constant rule states that if we differentiate a constant, the result is zero.
So let's differentiate the equation step-by-step:
dy/dt = b - 2ct
Therefore, the general expression for the rocket's velocity as a function of time is v(t) = b - 2ct.
(b) To find when the velocity is zero, we can set v(t) = 0 and solve for t:
0 = b - 2ct
Rearranging the equation, we get:
2ct = b
Dividing both sides by 2c gives us:
t = b / (2c)
So, the velocity of the rocket is zero at t = b / (2c).
To find the velocity of the rocket as a function of time, we need to find the derivative of the altitude equation, y = bt - ct^2, with respect to time.
(a) Differentiating the equation y = bt - ct^2 with respect to time t using the power rule, we get:
dy/dt = b - 2ct
So, the general expression for the rocket's velocity as a function of time is given by:
v(t) = b - 2ct
(b) To find when the velocity is zero, we set v(t) = 0 and solve for t:
b - 2ct = 0
2ct = b
t = b / (2c)
Substituting the given values for b = 82 m/s and c = 4.9 m/s^2 into the equation, we have:
t = 82 / (2 * 4.9)
t = 8.4 seconds
Therefore, the velocity is zero at t = 8.4 seconds.