Question

2. The equation of motion of a particle is given by
s=2t^3 - 9t^2+ 12t + 1, where s and t are measured in cm
and sec. The time when the particle stops momentarily is
(a) 1 sec
(b) 2 sec
(c) 1 sec, 2 sec
(d) None of these​

Answer 1

Given: Equation of motion of a particle is given by  s = 2t^3 - 9t^2+ 12t + 1

To find: The time when the particle stops momentarily?

Solution:

• Now we have given the equation of motion of a particle as:

                s = 2t^3 - 9t^2+ 12t + 1

• Here s and t are measured in cm.

• Now differentiating it with respect to time, we get:

                ds/dt = 6t^2 - 18t + 12

• Equating ds/dt with zero, we get:

                6t^2 - 18t + 12 = 0

                t^2 - 3t + 2 = 0

                (t - 2)(t - 1) = 0

• So here t = 1,2

Answer:

                  So the option c is correct, the time when the particle stops momentarily is 1 or 2 seconds.