Respuesta :
Answer:
[tex](-16.0\ \hat i+0\ \hat j)\ \rm m/s.[/tex]
Explanation:
Given:
- x-component of acceleration of the cart, [tex]\rm a_x = 4.0\ m/s^2.[/tex]
- y-component of acceleration of the cart, [tex]\rm a_y = 2.0\ m/s^2.[/tex]
- x-component of initial velocity of the cart, [tex]\rm v_{0x} = 8.0\ m/s.[/tex]
- y-component of initial velocity of the cart, [tex]\rm v_{0y} = 12\ m/s.[/tex]
Let the x and y components of the final velocity of the cart when the cart reaches its greatest y coordinate be [tex]\rm v_x\ and \ v_y[/tex] respectively.
When the cart reaches its greatest y-coordinate then the y component of its final velocity should be zero, [tex]\rm v_y =0.\\[/tex]
Let the cart reaches its greatest y-coordinate at time t, then using the following equation:
[tex]\rm v_y = v_{0y}+a_yt\\0=12+2\times t\\\Righttarrow t=\dfrac{-12}{2}=-6\ s.[/tex]
At this time, the x component of the velocity of the particle is given by
[tex]\rm v_x = v_{0x}+a_xt=8.0+4.0\times (-6)=8.0-24.0=-16.0\ m/s.[/tex]
Thus, the velocity of the cart when it reaches its greatest y coordinate, in unit-vector notation, is given by
[tex]\vec v = v_x\hat i+v_y\hat j=(-16.0\ \hat i+0\ \hat j)\ \rm m/s.[/tex]
