An atomic particle moves in a linear motion according to the formula s = 19 - 14t2 + 4t3. Determine the particle's velocity at the end of 2.9 seconds.
velocity in m/s =
My Answer
S'= 0-28t+12t^2
S'(2.9) = 0-28(2.9)+12(
2.9)^2
= 0-81.2+100.92
=19.72
Can you please confirm this answer
velocity in m/s = 19.72
To determine the velocity of the particle at the end of 2.9 seconds, we need to find the derivative of the given position function, s = 19 - 14t^2 + 4t^3, with respect to time. Let's find the derivative, denoted as s':
s' = d/dt (19 - 14t^2 + 4t^3)
= 0 - 28t + 12t^2.
Now, substitute t = 2.9 into the derivative equation:
s'(2.9) = 0 - 28(2.9) + 12(2.9)^2
= 0 - 81.2 + 100.92
= 19.72.
Therefore, the velocity of the particle at the end of 2.9 seconds is 19.72 m/s.
Your answer of 19.72 m/s is correct. Well done!