The displacement (in meters) of a particle moving in a straight line is given by s=2t^3 where is measured in seconds. Find the average velocity of the particle over the time interval [10,13].
the average velocity is 798
What is the instantaneous velocity of the particle when t=10?
distance gone during seconds 10, 11, and 12 /3
s(13) = 4394
s(10) = 2000
distance = 2394
av speed = 2394/3 = 798 m/s
ds/dt = v = 6 t^2
v(10) = 6 (100) = 600 m/s
444m/s
Well, when t=10, we can find the instantaneous velocity of the particle by taking the derivative of the displacement equation. So, let me calculate that for you.
The derivative of s = 2t^3 with respect to t is s' = 6t^2.
Now let's substitute t=10 into the derivative equation:
s' = 6(10)^2
s' = 6(100)
s' = 600
So, when t=10, the instantaneous velocity of the particle is 600 m/s. I hope that answers your question!
To find the instantaneous velocity of the particle when t = 10, we need to take the derivative of the displacement function with respect to time (t).
Given that the displacement function is s = 2t^3, we can find the velocity function v(t) by taking the derivative of s with respect to t:
v(t) = d(s)/dt = d(2t^3)/dt
To differentiate the function 2t^3, we can use the power rule of differentiation, which states that for any constant k, the derivative of kt^n with respect to t is nkt^(n-1).
In this case, k = 2 and n = 3. Applying the power rule, we get:
v(t) = 2 * 3t^(3-1)
= 6t^2
Now, to find the instantaneous velocity when t = 10, we substitute t = 10 into the velocity function v(t):
v(10) = 6(10)^2
= 600
Therefore, the instantaneous velocity of the particle when t = 10 is 600.