Question

A particle moves in the xy plane, starting from the origin at t=0 with initial velocity having an x component of 20 m/s and a y component of -15 m/s. The particle experiences an acceleration in the x direction, given by a x= 4.0 m/s2.
a) Determine the total velocity vector at any time.

b) Calculate the velocity and speed of the particle at t = 5.0 sec.

Answer 1

Answer:

a. $\vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}$

b. 25m/s

Explanation:

Let t be the time.

The velocity in the x direction that is subjected to acceleration of 4 m/s2

$v_x = 20 + 4t$ m/s

The velocity in the y direction

$v_y = -15$ m/s

The total velocity at any time, which is the combination of both x and y vectors of velocity

$\vec{v} = \vec{v_x} + \vec{v_y} = (20 + 4t)\hat{x} - 15 \hat{y}$

b) at t = 5

$v^2 = v_x^2 + v_y^2 = (20 + 4t)^2 + 15^2 = (20 + 4*5)^2 + 15^2 = 400^2 + 225 = 625$

$v = \sqrt{625} = 25 m/s$