Question

Q.1- At time t=0.00s, a particle located at the origin in the x-y plane has a velocity given by

3.00m/s i + 2.00m/s j. Three seconds later, the velocity is 9.00m/s i - 7.00m/s j. Assume

that the particle had a constant acceleration during those three seconds.

1- What was the acceleration of the particle?

2-What is the position of the particle at t=3.00s?

Answer 1

(1) Initial velocity = = ( 3 + 2 ) m / s, and, fnitial velocity = = ( 9 - 7 ) m / s.

Time taken to change this velocity = t = 3 s.

Hence, acceleration = = ( - ) / t

or, = [ ( 9 - 7 ) - ( 3 + 2 ) m / s ] / 3 s

or, = [ { ( 9 - 3 ) + ( - 7 - 2 ) } / 3 ] m / s²

or, = [ ( 6 - 9 ) / 3 ] m / s²

or, = ( 2 - 3 ) m / s².

(2) Along x - axis :

Initial velocity = u_x = 3 m / s,

Time travelled = t = 3 s,

Acceleration = a_x = 2 m / s².

Hence, position along x - axis = x = u_xt + a_xt² / 2

or, x = [ 3 x 3 + 2 x 3² / 2 ] m

or, x = 18 m.

Along y - axis :

Initial velocity = u_y = 2 m / s,

Time travelled = t = 3 s,

Acceleration = a_y = - 3 m / s².

Hence, position along y - axis = x = u_yt + a_yt² / 2

or, y = [ 2 x 3 + ( - 3 ) x 3² / 2 ] m

or, y = - 7.5 m.

Hence, position of the particle after three seconds is : ( 18 - 7.5 ) m.