A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed when t = 5. (Round your answers to two decimal places.)

f(t) = t^−1 − t

*We havent learned derivative yet but we have learned difference quotient and limits

Damon, where did you get the -h from?

Damon, also how did you get f(t)=1/t -1? Shouldn't f(t)=1/t-t

Well, it looks like we have a mathematical challenge here! Since we haven't learned derivatives yet, let's use the difference quotient to find the velocity.

The velocity of an object is defined as the derivative of its position with respect to time. In this case, the position function is given by f(t) = t^(-1) - t.

To find the velocity, we can use the difference quotient formula:

v = lim(h->0) [f(t+h) - f(t)] / h

Let's calculate it step by step:

f(t+h) = (t+h)^(-1) - (t+h)
f(t) = t^(-1) - t

Now, let's substitute these values into the difference quotient formula:

v = lim(h->0) [(t+h)^(-1) - (t+h) - (t^(-1) - t)] / h

Simplifying,

v = lim(h->0) [(t+h)^(-1) - t^(-1)] / h

Now, we substitute t = 5 into this expression:

v = lim(h->0) [(5+h)^(-1) - 5^(-1)] / h

To calculate this limit, you can use the tools you have learned for limits and difference quotients. After finding the limit, you'll get the velocity at t = 5.

As for the speed, it is the absolute value of the velocity. Once you have the velocity, simply take its absolute value to find the speed.

Now, go ahead and give it a try! Good luck! And remember, math problems can be a bit tricky, so keep your sense of humor close by!

To find the velocity when t = 5, we can calculate the average velocity over a small interval surrounding t = 5. We will use the difference quotient to do this.

First, let's calculate the average velocity over an interval around t = 5. We will use a small interval of Δt = 0.01 seconds. The average velocity is given by the formula:

Average velocity = (f(5+Δt) - f(5)) / Δt

Substituting the given equation of motion, f(t) = t^(-1) - t:

Average velocity = ((5+Δt)^(-1) - (5+Δt)) - (5^(-1) - 5) / Δt

Next, we can simplify the expression by expanding and combining like terms:

Average velocity = ((5+Δt)^(-1) - (5+Δt)) - (1/5 - 5) / Δt

Note: The difference quotient is used to calculate the average rate of change, which is a close approximation of the instantaneous rate of change (i.e., velocity) as the interval Δt approaches zero. In this case, we are using a small Δt value to approximate the velocity at t = 5.

Now, we will calculate the average velocity by plugging in Δt = 0.01 into the expression above:

Average velocity = ((5+0.01)^(-1) - (5+0.01)) - (1/5 - 5) / 0.01

Simplifying further:

Average velocity = (1/5.01 - 5.01) - (-0.8) / 0.01

Average velocity ≈ -0.201

Therefore, the velocity of the particle when t = 5 is approximately -0.201 m/s (rounded to two decimal places).

To calculate the speed when t = 5, we need to take the magnitude of the velocity. Since the velocity is negative, the speed will be positive. Hence, the speed is approximately 0.201 m/s (rounded to two decimal places).

If all in one direction, the velocity is the speed.

To get speed you need a speedometer.
To get velocity you need a speedometer and a compass.

If you knew derivatives this would be easy
v = df/dt = -1 (t^-2) - 1
= -1/t^2 - 1
for t = 5
v = -1/25 -25/25 = -26/25
--------------------------
if not then a mess as follows
f(t+h) = 1/(t+h) -t -h
f(t) = 1/t - t

f(t+h) - f(t) = 1/(t+h)- 1/t -h
= [ t -t - h ]/[t(t+h)] - h
divide by h
[ -1 ]/[t(t+h)] - 1
let h---> 0
-1/t^2 -1 like we said