The displacement (in meters) of a particle moving in a straight line is given by the equation of motion s = 4/t^2, where t is measured in seconds. Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

ds/dt = v = -4 (2t) /t^4 = -8/t^3

if t = a
v = -8/a^3

if t = 1
v = -8/1 = -8

if t = 2
v = -8/8 = -1

if t = 3
v = -8/27

Well, well, well! Looks like we have a particle on the move. Let's get crackin' on finding its velocity at different times.

To find the velocity, we need to first find the derivative of the displacement equation. So, let's put on our math caps and dive in!

Given that s = 4/t^2, we need to differentiate it with respect to time (t):

ds/dt = d(4/t^2)/dt

Now, my trusty mathematical friend, the chain rule, comes in handy here. Hang tight!

ds/dt = -8/t^3

Voila! We've got the derivative of the displacement equation. Now, let's plug in different values of t and have some fun finding the velocities.

At t = a, we have ds/dt = -8/a^3. But wait, what's the value of a? You didn't provide it, my dear interlocutor. So, I'm afraid I can't specifically calculate the velocity at t = a without knowing the numerical value for a.

However, let's have a blast finding the velocity at t = 1, t = 2, and t = 3.

For t = 1, the velocity v1 = ds/dt = -8/1^3 = -8 m/s. Negative, but still moving!

For t = 2, the velocity v2 = ds/dt = -8/2^3 = -1 m/s. The particle appears to be slowing down.

For t = 3, the velocity v3 = ds/dt = -8/3^3 = -8/27 m/s. The particle is definitely losing steam.

So, my friend, we have the velocities at t = 1, 2, and 3. Unfortunately, I couldn't calculate it for t = a, but hey, the show must go on!

To find the velocity of the particle at different times, we need to find the derivative of the equation of motion with respect to time.

Taking the derivative of s = 4/t^2 with respect to t gives:

ds/dt = d(4/t^2)/dt

To differentiate this expression, we can use the power rule for differentiation:

ds/dt = -8/t^3

Now, let's substitute the given values of time (a, 1, 2, and 3) into this derivative expression to find the velocities at those times.

At t = a:
v(a) = ds/dt = -8/a^3

At t = 1:
v(1) = ds/dt = -8/1^3 = -8

At t = 2:
v(2) = ds/dt = -8/2^3 = -1

At t = 3:
v(3) = ds/dt = -8/3^3 = -8/27

Therefore, the velocities of the particle at times t = a, t = 1, t = 2, and t = 3 are -8/a^3, -8, -1, and -8/27 respectively.

To find the velocity of the particle at different times, we need to calculate the derivative of the displacement equation, which gives us the rate of change of displacement with respect to time. In other words, the derivative gives us the velocity function.

Given the equation of motion s = 4/t^2, we can find the velocity function by taking the derivative with respect to time (t):

ds/dt = d(4/t^2)/dt

To find the derivative, we can use the power rule: d(x^n)/dx = n*x^(n-1)

So, applying the power rule to the equation, we get:

ds/dt = -8/t^3

Now that we have the velocity function, we can substitute different values of time (t) to find the velocity at those times.

Velocity when t = a:
v(a) = -8/a^3

Velocity when t = 1:
v(1) = -8/1^3 = -8

Velocity when t = 2:
v(2) = -8/2^3 = -1

Velocity when t = 3:
v(3) = -8/3^3 = -8/27

Therefore, the velocity of the particle at times t = a, t = 1, t = 2, and t = 3 are -8/a^3, -8, -1, and -8/27 meters per second, respectively.