(a) Expression for the velocity is -
v(t) = (6t - 3t^2) m/s
To find out the acceleration, take differentiation of this -
Acceleration, a(t) = dv(t)/dt = (6 - 2*3t) m/s^2 = (6 - 6t) m/s^2
Put t = 3 s
a(t) = (6 - 6*3) = -12 m/s^2
Negative sign shows it is deceleration.
Magnitude of deceleration = 12 m/s^2
(b) To find out the distance, integrate the expression of velocity -
x(t) = 6*t^2/2 - 3*t^3/3 + k = 3*t^2 - t^3 + k (where k is a constant)
at, t = 0, x = 0
so, we have -
k = 0
So our expression becomes -
x(t) = 3*t^2 - t^3
put t = 3 s
x(3) = 3*3^2 - 3^3 = 0 meter
So, distance traveled by the particle during 3 s = 0
Average speed = distance traveled/time taken = 0/3 = 0 m/s
Please show all work and answer parts a and b! Thank you! Problem 1: The velocity...
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