The equation of motion of a particle is s=4t^3−7t, where s is in meters and t is in seconds. Find
a) the velocity of the particle as a function of t : v(t)=
b) the acceleration of the particle as a function of t : a(t)=
c) the velocity after 5 seconds
d) the acceleration after 5 seconds
Find a cubic function y=ax3^+bx^2+cx+d whose graph has horizontal tangents at the points (-1, -16) and (1,12).
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Patrick's problem:
v(t) = 12t^2 - 7
a(t) = 24t
c) v(5) = 12(25) - 7 = ...
d) a(5) = .....
To find the velocity and acceleration of the particle, we need to differentiate the equation of motion with respect to time.
a) Velocity (v(t)) can be found by taking the first derivative of the equation of motion.
s = 4t^3 - 7t
To find v(t), differentiate s with respect to t:
v(t) = d/dt (4t^3 - 7t)
Using the power rule of differentiation, we get:
v(t) = 12t^2 - 7
Therefore, the velocity of the particle as a function of t is v(t) = 12t^2 - 7.
b) Acceleration (a(t)) can be found by taking the second derivative of the equation of motion.
To find a(t), differentiate v(t) with respect to t:
a(t) = d/dt (12t^2 - 7)
Using the power rule of differentiation, we get:
a(t) = 24t
Therefore, the acceleration of the particle as a function of t is a(t) = 24t.
c) To find the velocity after 5 seconds (v(5)), substitute t = 5 into the velocity equation:
v(5) = 12(5)^2 - 7
v(5) = 12(25) - 7
v(5) = 300 - 7
v(5) = 293
Therefore, the velocity of the particle after 5 seconds is 293 m/s.
d) To find the acceleration after 5 seconds (a(5)), substitute t = 5 into the acceleration equation:
a(5) = 24(5)
a(5) = 120
Therefore, the acceleration of the particle after 5 seconds is 120 m/s^2.