Open Question
(II) At t = 0, a particle starts from rest at 𝓍 = 0 , y = 0 and moves in the xy plane with an acceleration a (→ above a) = (4.0 î + 3.0 ĵ ) m/s². Determine
(b) the speed of the particle as a function of time,
4. 2D Kinematics
Kinematics in 2D
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A survey drone has just completed a scan at x,y coordinates (57m, 8m) at t=0. It needs to return to a lab located at (-115, 72) m. If its initial velocity is 16m/s in the +y-direction, and it has only 18s of battery life remaining, what constant acceleration (magnitude and direction) does it need to reach the lab?
A
2.8 m/s2; along –x axis
B
1.8 m/s2; 51.8° below –x axis
C
2.8 m/s2; above –x axis
D
1.3 m/s2; 24° above –x axis
Verified step by step guidance1
Identify the initial and final positions of the drone. The initial position is (57 m, 8 m) and the final position is (-115 m, 72 m).
Calculate the displacement vector \( \Delta \mathbf{r} \) by subtracting the initial position from the final position: \( \Delta x = -115 - 57 \) and \( \Delta y = 72 - 8 \).
Determine the initial velocity vector \( \mathbf{v}_0 \). Given that the initial velocity is 16 m/s in the +y-direction, the vector is \( \mathbf{v}_0 = (0, 16) \) m/s.
Use the kinematic equation for constant acceleration: \( \Delta \mathbf{r} = \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \). Substitute the known values for \( \Delta \mathbf{r} \), \( \mathbf{v}_0 \), and \( t = 18 \) s to solve for the acceleration vector \( \mathbf{a} \).
Calculate the magnitude and direction of the acceleration vector \( \mathbf{a} \). The magnitude is given by \( |\mathbf{a}| = \sqrt{a_x^2 + a_y^2} \) and the direction \( \theta \) can be found using \( \tan(\theta) = \frac{a_y}{a_x} \).
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