A science project studying catapults sent a projectile into the air with an initial velocity of 30 m/s. The formula for distance (s) in meters with respect to time (t) in seconds is s(t) = -5t2 + 30t. Find an expression for the instantaneous rate of change (and thus a velocity function at any time) using s'(a)=limh→0s
((a+h)−s(a))/h where a is any value for t. Then, find the instantaneous rate of change at: a) t = 1 b) t = 3
s(t) = -5 t^2 + 30 t we want the time derivative which is velocity
s(t+h) = -5 (t+h)^2 + 30 (t+h)
= -5 (t^2 + 2 t h + h^2) + 30 t + 30 h
= -5 t^2 - 10 t h - 5 h^2 + 30 t + 30 h
now subtract s(t) = - 5 t^2 + 30 t
result = -10 t h -5 h^2 + 30 h
now divide by h
result = -10 t - 5 h + 30
now let h -----> 0
derivative = velocity = -10 t + 30
note that is if you shoot it straight up at 30 meters/second and g = -10 meters/second^2
now I think you can put in t = 1 and t = 3
note --- t = 3 seconds is the tippy top, when it stops and then falls
To find the instantaneous rate of change, we first need to find the derivative of the distance function.
Given the distance function s(t) = -5t^2 + 30t, we can find the derivative by differentiating each term separately with respect to t.
s'(t) = d/dt(-5t^2 + 30t)
= -10t + 30
Now, we can find the instantaneous rate of change by evaluating s'(a).
s'(a) = -10a + 30
To find the instantaneous rate of change at t = 1, we substitute a = 1 into the expression for s'(a).
s'(1) = -10(1) + 30
= -10 + 30
= 20
Therefore, the instantaneous rate of change at t = 1 is 20 m/s.
To find the instantaneous rate of change at t = 3, we substitute a = 3 into the expression for s'(a).
s'(3) = -10(3) + 30
= -30 + 30
= 0
Therefore, the instantaneous rate of change at t = 3 is 0 m/s.
To find the expression for the instantaneous rate of change or the velocity function at any time, we need to find the derivative of the function s(t) = -5t^2 + 30t with respect to time (t).
The derivative of a function represents the rate at which the function is changing at any given point. In this case, we want to find the derivative of s(t) to get the rate of change or velocity at any time.
Step 1: Find the derivative of s(t)
s(t) = -5t^2 + 30t
To find the derivative, we can use the power rule of differentiation. The power rule states that if we have a term of the form x^a, then its derivative is a * x^(a-1).
Applying the power rule to each term in s(t), we get:
s'(t) = -5 * 2t^(2-1) + 30 * 1t^(1-1)
= -10t + 30
So the derivative or the instantaneous rate of change function is s'(t) = -10t + 30.
Step 2: Find the instantaneous rate of change at specific values of t.
a) To find the instantaneous rate of change at t = 1, we substitute t = 1 into the expression for s'(t).
s'(1) = -10 * 1 + 30
= -10 + 30
= 20
Therefore, the instantaneous rate of change or velocity at t = 1 is 20 m/s.
b) To find the instantaneous rate of change at t = 3, we substitute t = 3 into the expression for s'(t).
s'(3) = -10 * 3 + 30
= -30 + 30
= 0
Therefore, the instantaneous rate of change or velocity at t = 3 is 0 m/s.