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Chapter 2: Q. 36 (page 61)

The position of a particle is given by the function x=2t3-6t2+12m, where t is in s. a. At what time does the particle reach its minimum velocity? What is Vxmin? b. At what time is the acceleration zero?

Short Answer

Expert verified

(a)The particle reaches its minimum velocity at t=1s. The minimum velocity of the particle is -6ms.

(b) The acceleration of the particle is zero att=1s.

Step by step solution

01

Given information

The position of the particle is given byx=2t3-6t2+12
.

02

The velocity of the particle.

The velocity is given by vx=dxdt.

Differentiate the given function with respect to t.

vx=ddt2t3-6t2+12=6t2-12t

03

Acceleration of the particle.

To find the acceleration, differentiate vx=6t2-12twith respect to t.

ax=ddt6t2-12t=12t-12

Substitute 0 for ax.

0=12t-1212t=12t=1s

04

Part (a)

The velocity of the particle at t=1sis

vx=612-121=6-12=-6ms

The particle reaches its minimum velocity at t=1s. The minimum velocity of the particle is -6ms.

05

Part (b)

The acceleration of the particle is zero att=1s.

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