Question

A particle moves in the x-y plane with a constant acceleration of 8 m/s2 making an angle 37° with x-axis. At t=0 the particle is at the origin and its velocity is 8 m/s along x-axis. Find velocity and the position of the particle at t = 2 s. ​

Answer 1

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north).

Explanation:

September 27 of 1996

refer to the attachment

Answer 2

Answer:

Net Velocity of the particle at t = 2 s  is 22.88 m/s  and Final position of the particle at t = 2 s  is  ( 28.76 , 9.62 ).

Explanation:

Given,

$u_{x}$ = 8 m/s

a = 8 m/$s^{2}$

Acceleration of particles along x-axis  = $a_{x}$  =  a × cos 37°

                                                                        = 8 cos 37°  m/$s^{2}$

                                                                        = 6.38 m/$s^{2}$

Acceleration of particles along y-axis = $a_{y}$  =  a × sin 37°

                                                                       = 8 sin 37°  m/$s^{2}$    

                                                                       = 4.81 m/$s^{2}$

After 2 sec, Velocity of particles in x-axis = $V_{x}$  =  $V_{0} + a_{x}t$

                                                                             = 8 m/s + 6.38 m/$s^{2}$ × 2 sec

                                                                             = 20.76 m/s

After 2 sec, Velocity of particles in y-axis = $V_{y}$  =  $a_{y} t$

                                                                             = 4.81 m/$s^{2}$ × 2 sec

                                                                             = 9.62 m/s

Hence,

Net Velocity of the particle at t=2 s   = $\sqrt{V_{x}^{2} +V_{y}^{2} }$

                                                             = $\sqrt{(20.76)^{2} + (9.62)^{2} }$

                                                             = $\sqrt{430.98 + 92.54}$

                                                             = 22.88 m/s

∴ Net Velocity of the particle at t = 2 s  is 22.88 m/s

Now,

Position of particle:

x  = $u_{x} t + \frac{1}{2} a_{x} t^{2}$                                             y  =    $u_{y} t + \frac{1}{2} a_{y} t^{2}$

   = $8(2) + \frac{1}{2} (6.38)(2)^{2}$                                       =  $0 (2) + \frac{1}{2} (4.81)(2)^{2}$

   = 28.76                                                         = 6.62

∴ Final position of the particle at t = 2 s  is  ( 28.76 , 9.62 ).