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2 years ago

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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- p

onents 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?

1 answer:

Tasya [4]2 years ago

4 0

Answer:

$(-16.0\ \hat i+0\ \hat j)\ \rm m/s.$

Explanation:

<u>Given:</u>

x-component of acceleration of the cart, $\rm a_x = 4.0\ m/s^2.$

y-component of acceleration of the cart, $\rm a_y = 2.0\ m/s^2.$

x-component of initial velocity of the cart, $\rm v_{0x} = 8.0\ m/s.$

y-component of initial velocity of the cart, $\rm v_{0y} = 12\ m/s.$

Let the x and y components of the final velocity of the cart when the cart reaches its greatest y coordinate be $\rm v_x\ and \ v_y$ respectively.

When the cart reaches its greatest y-coordinate then the y component of its final velocity should be zero, $\rm v_y =0.\\$

Let the cart reaches its greatest y-coordinate at time t, then using the following equation:

$\rm v_y = v_{0y}+a_yt\\0=12+2\times t\\\Righttarrow t=\dfrac{-12}{2}=-6\ s.$

At this time, the x component of the velocity of the particle is given by

$\rm v_x = v_{0x}+a_xt=8.0+4.0\times (-6)=8.0-24.0=-16.0\ m/s.$

Thus, the velocity of the cart when it reaches its greatest y coordinate, in unit-vector notation, is given by

$\vec v = v_x\hat i+v_y\hat j=(-16.0\ \hat i+0\ \hat j)\ \rm m/s.$

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