A rocket is fired into the air with an initial velocity of 98 m/s. The distance (s) the rocket is above the ground, in metres, after t seconds is given by the expression s(t) = 98t - 4.9t2. What is the rocket's instantaneous rate of change at the moment it hits the ground? Include proper units and a direction.
s = 98 t - 4.9 t^2
differentiate to get velocity
v = 98 - 9.8 t
now
solve that top equation for t with s = 0 at ground
4.9 t^2 - 98 t = 0
t (4.9 t -98) = 0
obviously the height is zero when t = 0
but also when t = 98/4.9 = 20
so v = 98 - 9.8*20
= - 98 m/s
which we could have said by symmetry. It has the same speed leaving and hitting, but sign changes.
To find the rocket's instantaneous rate of change at the moment it hits the ground, we need to find the derivative of the function s(t).
The given function is s(t) = 98t - 4.9t^2.
To find the derivative, we differentiate the function with respect to t:
s'(t) = d/dt (98t - 4.9t^2)
To differentiate, we use the power rule for differentiation. For a term of the form ax^n, the derivative is nax^(n-1).
Differentiating each term in the expression, we get:
s'(t) = 98 - 9.8t
The rocket hits the ground at the moment when s(t) = 0. Therefore, we set s(t) = 0 and solve for t:
0 = 98t - 4.9t^2
Rearranging the equation, we get:
4.9t^2 - 98t = 0
Factoring out t, we have:
t(4.9t - 98) = 0
Setting each factor equal to zero, we find two solutions:
t = 0 (this is the initial time, when the rocket is fired)
4.9t - 98 = 0
Solving the second equation, we find:
4.9t = 98
t = 98 / 4.9
t = 20
So, the rocket hits the ground after 20 seconds.
To find the rocket's instantaneous rate of change at this moment, we substitute t = 20 into the derivative equation we found earlier:
s'(20) = 98 - 9.8(20)
= 98 - 196
= -98 m/s
The minus sign indicates that the rate of change is negative, which means the rocket is moving downward. Therefore, the rocket's instantaneous rate of change at the moment it hits the ground is -98 m/s, in the downward direction.