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A particle leaves the origin with a velocity of 7.2 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i-3.0j) m/s^2. At the instant the particle moves back across the x axis (y= 0), what is the value of its x coordinate?

Sagot :

The origin is (0,0)

The initial velocity

[tex]\vec{v_i}=7.2\text{ m/s }\hat{\text{j}}[/tex]

Acceleration is constant

[tex]\vec{a}=\mleft(2.0\hat{i}-3.0\hat{j}\mright)\text{ m/s}^2[/tex]

When the particle again reaches x-axis the position is (x,0)

[tex]a_y=-3j\text{m/s}^2[/tex][tex]y=0[/tex]

time taken =t

[tex](v_i)_y=7.2j\text{ m/s}[/tex][tex]y=v_{yi}t+\frac{1}{2}a_yt^2_{}[/tex][tex]\begin{gathered} 0=(7.2)t+\frac{1}{2}(-3)t^2 \\ t=4.8\text{ s} \end{gathered}[/tex]

Now for the coordinate x

[tex]v_{x\text{i}}=0\text{ m/s}[/tex][tex]a_x=2m/s^2[/tex][tex]x=0+\frac{1}{2}(2)(4.8)^2=23.04m[/tex]

ANSWER

x=23.04m