A particle starts from the origin at t = 0 with a velocity of (16 i? -12 j? )m/s and moves in the xy plane with a constant acceleration of $\overrightarrow { \mathbf { a } }$ = (3.0 i? - 6.0 j? ) m/s². What is the speed of the particle at t = 2.0 s? A)52 m/s B)39 m/s C)46 m/s D)33 m/s E)43 m/s

Question

Quizplus · Accepted Answer

The velocity at t = 2.0 s is given by v = u + at, where u is the initial velocity and a is the acceleration. Substituting the given values, v = (16 i - 12 j) + 2*(3 i - 6 j) = (16 + 6) i + (-12 - 12) j = 22 i - 24 j. The speed is the magnitude of the velocity vector: √(22^2 + (-24)^2) = √(484 + 576) = √1060 = 33 m/s.

A Particle Starts from the Origin at T = 0 a→\overrightarrow { \mathbf { a } }a

A Particle Starts from the Origin at T = 0 $\overrightarrow { \mathbf { a } }$