A cart is propelled over an xy plane with acceleration compo- nents ax 4.0 m/s2 and ay 2.0 m/s2. Its initial velocity has com- ponents v0x 8.0 m/s and v0y 12 m/s. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?

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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]

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