The position of a particle, s metres from the starting point, after t seconds, is defined by the function s(t) = 2t^3 - 7t^2 + 4
a) determine the velocity of the particle after t seconds
b) determine the velocity after 5 seconds
v(t) = s'(t) = 6t^2-14t
v(5) = 6(5)^2 - 14(5)
= 150-70 = 80
thank you!
To determine the velocity of the particle, we need to find the derivative of the position function with respect to time, which will give us the rate of change of position with respect to time.
a) To find the velocity, we need to take the derivative of the position function s(t) = 2t^3 - 7t^2 + 4 with respect to time (t).
Step 1: Differentiate each term of the function separately using the power rule of differentiation.
ds/dt = d(2t^3)/dt - d(7t^2)/dt + d(4)/dt.
Step 2: Apply the power rule of differentiation:
- For the term 2t^3, the derivative is 6t^2.
- For the term -7t^2, the derivative is -14t.
- For the constant term 4, the derivative is 0 since a constant has a derivative of zero.
Therefore, the velocity equation v(t) = ds/dt is:
v(t) = 6t^2 - 14t + 0,
v(t) = 6t^2 - 14t.
b) To determine the velocity after 5 seconds, substitute t = 5 into the velocity equation v(t) = 6t^2 - 14t.
v(5) = 6(5)^2 - 14(5),
v(5) = 6(25) - 70,
v(5) = 150 - 70,
v(5) = 80.
Therefore, the velocity after 5 seconds is 80 m/s.