An object is observed moving along a straight line parallel to the x axis. The s

Question

An object is observed moving along a straight line parallel to the x axis. The scalar x component of the object's vel is given as a function of time by (32 -18t) ,where t is the number of seconds since observations began. At t = 0, the object's x-position coordinate is equal to + 2 m. 1. Att 2 seconds, the object is located at x equals a. 28 m. b.+28 m c.-26 m e.-24 m 2. When, if ever, is the object moving in the negative x direction? a. t26s 3. Att-2 seconds, the object is a. slowing down while moving in the positive x direction. b. slowing down while moving in the negative x direction. c. speeding up while moving in the positive x direction. d. speeding up while moving in the negative x direction. e. at rest.

Explanation / Answer

1) Vx = dx/dt = 3t2 - 18 t
dx = (3t2 - 18 t)dt
integrating with limits on LHS from +2 to X (position of particle at t=2 sec)and on right side from 0 to 2, we get
X-2 = (23 - 9*22) = -28
X = -26

2) For object to move in -ve direction, vx should have -ve sign
3t2 - 18 t < 0
3t2 < 18 t
considering +ve values of t,
t < 6 (option D is correct)

3) at 2 sec, Vx = 3t2 - 18 t = -24 m/s
acceleration ax = dVx/dt = 6t - 18
at t= 2 , ax = -6

As a and v have same sign , particle is speeding up.
Option d is correct.

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