A model rocket is launched straight upward. Its altitude y as a function of time is given by y=bt−ct2, where b = 80m/s , c = 4.9m/s2 , t is the time in seconds, and y is in meters.
1. Use differentiation to find a general expression for the rocket's velocity as a function of time.
2. When is the velocity zero?
Express your answer using two significant figures.
v = dy/dt = b - 2ct
now just solve for t when v=0
1. Well, let's put on our differentiation hat and find the rocket's velocity. We need to differentiate the altitude function, y, with respect to time, t.
So, dy/dt = d(b t - c t^2)/dt
Now, the derivative of a constant times t is just the constant, so the first term gives us b.
The derivative of -c t^2 is -2c t.
Therefore, the velocity, v, is given by v = dy/dt = b - 2c t.
So, the general expression for the rocket's velocity as a function of time is v = 80 - 2(4.9)t.
2. To find when the velocity is zero, we set v = 0 and solve for t:
0 = 80 - 2(4.9)t
Simplifying this, we get:
2(4.9)t = 80
9.8t = 80
t ≈ 8.16 seconds (rounded to two significant figures).
So, the velocity of the rocket is zero at approximately 8.16 seconds.
To find the velocity of the rocket as a function of time, we need to differentiate the equation for altitude with respect to time.
Given: y = bt - ct^2
To find the velocity, we differentiate y with respect to t:
v = d/dt (bt - ct^2)
Differentiating each term separately, we have:
v = d/dt (bt) - d/dt (ct^2)
Since b and c are constants, their derivatives will be zero:
v = b - 2ct
Therefore, the general expression for the rocket's velocity as a function of time is:
v = b - 2ct
Now, to find when the velocity is zero, we set v = 0 and solve for t:
0 = b - 2ct
Rearranging the equation, we have:
2ct = b
Solving for t, we get:
t = b / 2c
Substituting the given values b = 80 m/s and c = 4.9 m/s^2, we can calculate the value of t:
t = 80 / (2 * 4.9) = 8.16 seconds (rounded to two significant figures)
Therefore, the velocity of the rocket is zero at approximately 8.16 seconds.
To find the rocket's velocity as a function of time, we need to take the derivative of the given altitude function with respect to time.
1. Differentiating the altitude function, y = bt - ct^2, with respect to time (t), we can use the power rule of differentiation. The power rule states that if we have a monomial of the form ax^n, the derivative will be nx^(n-1).
Differentiating bt with respect to t: b * (t^1) = bt
Differentiating -ct^2 with respect to t: -2c * (t^1) = -2ct
The derivative of y with respect to t is given by: dy/dt = bt - 2ct
So, the general expression for the rocket's velocity as a function of time is v(t) = bt - 2ct.
2. To find when the velocity is zero, we need to set the velocity function equal to zero and solve for t.
Setting v(t) = 0:
bt - 2ct = 0
Factoring out t:
t(b - 2c) = 0
From this equation, we have two possibilities for when the velocity is zero:
1. When t = 0 (the initial time when the rocket is launched).
2. When b - 2c = 0.
To find the second possibility, we need to solve for t:
b - 2c = 0
2c = b
t = b / (2c)
Using the given values:
b = 80 m/s
c = 4.9 m/s^2
Substituting these values into the equation, we can solve for t:
t = 80 / (2 * 4.9)
t ≈ 8.16 seconds
Therefore, the velocity is zero at approximately 8.16 seconds.