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 in seconds is
s = -4.9t2 + 30t. Find the instantaneous rate of change (and thus the velocity at any time) using and then find the instantaneous rate of change at:
t = 1
t = 3
That formula applies only if the projectile is launched straight up. That would not be a practical way of launching an object with a catapult, unless you are doing a Monty Python movie and launching dead cows.
Unless the motion is in a straight line, as it is in this case, the rate of change of the distance away, ds/dt, is NOT the velocity.
Having said that, the rate of change of s is 30 - 9.8t m/s
Plug in t = 1 and then t=3 for the velocities they ask for.
To find the instantaneous rate of change (velocity) at any given time, we need to take the derivative of the distance equation with respect to time.
The derivative of the equation s = -4.9t^2 + 30t is given by:
ds/dt = -9.8t + 30
Now we can substitute the values of t to find the instantaneous rate of change at t = 1 and t = 3.
At t = 1:
ds/dt = -9.8(1) + 30
= -9.8 + 30
= 20.2 m/s
Therefore, the instantaneous rate of change (velocity) at t = 1 is 20.2 m/s.
At t = 3:
ds/dt = -9.8(3) + 30
= -29.4 + 30
= 0.6 m/s
Therefore, the instantaneous rate of change (velocity) at t = 3 is 0.6 m/s.
To find the instantaneous rate of change (velocity) at any given time, we need to calculate the derivative of the distance formula with respect to time. The derivative represents the rate of change of the distance function.
Given the formula for distance:
s = -4.9t^2 + 30t
To find the derivative, we differentiate each term separately using the power rule, product rule, and sum rule of derivatives.
1. Differentiating the first term:
d/dt (-4.9t^2) = -9.8t
2. Differentiating the second term:
d/dt (30t) = 30
Therefore, the derivative of the distance function, which represents the instantaneous rate of change (velocity) at any given time, is:
v(t) = -9.8t + 30
Now, we can find the instantaneous rate of change (velocity) at t = 1 and t = 3.
1. At t = 1:
Substitute t = 1 into the velocity function:
v(1) = -9.8(1) + 30
v(1) = -9.8 + 30
v(1) = 20.2
So, the instantaneous rate of change (velocity) at t = 1 is 20.2 m/s.
2. At t = 3:
Substitute t = 3 into the velocity function:
v(3) = -9.8(3) + 30
v(3) = -29.4 + 30
v(3) = 0.6
So, the instantaneous rate of change (velocity) at t = 3 is 0.6 m/s.