Question

If an object has an initial position
(x1, y1) = (5.0 m, 9.0 m),
and an initial velocity
(v1x, v1y) = (2.0 m/s, −9.0 m/s),
what is the final position vector
(x2, y2)
of the object if its two-dimensional acceleration is described by the graphs in the figure below? (Express your answer in vector form in metres.)

Answer 1

Answer on Question #62290-Physics-Atomic and Nuclear Physics

If an object has an initial position (x1, y1) = (5.0 m, 9.0 m), and an initial velocity (v1x, v1y) = (2.0 m/s, -9.0 m/s), what is the final position vector (x2, y2) of the object if its two-dimensional acceleration is described by the graphs in the figure below? (Express your answer in vector form in metres.)

Solution

$$\boldsymbol{a} = (a_x, a_y)$$ $$\boldsymbol{p}_f = \boldsymbol{p}_i + \boldsymbol{v}_i t + \boldsymbol{a} \frac{t^2}{2}$$ $$\boldsymbol{p}_f = (5.0 \text{ m}, 9.0 \text{ m}) + \left(2.0 \frac{\text{m}}{\text{s}}, -9.0 \frac{\text{m}}{\text{s}}\right) t + (a_x, a_y) \frac{t^2}{2}.$$

If you have the graphs in the figure then you can paste the values $a_x, a_y, t$ in this formula to obtain the answer in only numbers.

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