Question

Please show all work and answer parts a and b! Thank you!

Problem 1: The velocity of a particle traveling in a straight line is given by v- (6t 3t) m/s, where t is in seconds. If x-0 when t = 0, (a) determine the particle's deceleration and position when t = 3s. b) How far has the particle traveled during the 3-s time interval, and what is its average speed?

Answer 1

(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