A wrench released from a state of rest at time 𝑡=0 travels a distance 𝑠(𝑡)=4.9𝑡^2 m in 𝑡 seconds. Estimate the instantaneous velocity at 𝑡=9 seconds. 𝑣= m/s

Answer 1

To be clear, "A wrench released from a state of rest" just means the wrench was dropped and is falling straight down. So our function s(t) represents how far the wrench has fallen at any time t (the number 4.9 comes from the strength of gravity).

Now the problem is asking us to find the instantaneous velocity. To solve this, you must know that velocity is the derivative of position, because velocity is defined as the change in position over time. Consider the following:

v = (p2 - p1) / (t2 - t1) -> This is the formula for velocity where p1 is the start

position at time t1, and p2 is the end position at time t2.

In two steps, this will turn into the definition of a derivative:

1. We can take position to be a function of time, so p2 = f(t2) and p1 = f(t1)

2. We can take t1 to be a starting point and represent t2 as some duration h after

t1, like t2 = t1 + h

Now, if we rewrite the formula for speed using these substitutions, we get:

v = ( f(t1 + h) - f(t1) ) / ( (t1 + h) - t1) -> This is precisely the definition

of the derivative.

So now that we know velocity is the derivative of position, we can simply take the derivative of our function s(t) (which is position as a function of time) to get s'(t) = v(t) (our velocity as a function of time). Once we figure out the derivative, it's only a matter of plugging in t = 9:

s(t) = 4.9t^2 -> Position as a function of time

s'(t) = v(t) = 9.8t -> Derivative of position or velocity as a function of time

s'(9) = v(9) = 9.8(9) = 88.4 -> Velocity at t = 9

Last thing, don't forget the units to your answer! It states that velocity is in meters per second, so your final answer should be 88.4m/s.

Answer 2

s(t) = -4.9t^2

velocity = the derivative

s'(t) = -9.8t

s'(9) = -9.8(9) = -88.2 meters per second = velocity at 9 seconds