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
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
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!
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.
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.