Question

The ball X has a known position at 3 different times, each 1 second apart. The positions of the ball are (1.8,2.2)m, (4.4,4.8)m, (6.4,7.9)m, for times t-1, t, and t+1 respectively. What is the approximate magnitude of the accelerate of the ball at time t (in m/s^2), given this information?

Answer 1

Answer:

Explanation:

From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.

Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m

So velocity along x direction V_x = $\frac{2.6}{s}$

Similarly velocity along y direction V_y(1) =  $\frac{2.6}{s}$

In the next phase velocity changes both in x and y direction.

velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex

Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex

Acceleration in x direction = change of velocity in x direction

= ( 2 - 2.6 ) = -.6 m s⁻²

Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²

Total acceleration =\sqrt{( -.6 )² + ( .5 )²}

= .78 ms⁻²