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Additional Problems (2D kinematics) Example 1 A rabbit runs across a parking l...
May 30, 2024
Solution
a
To find the rabbit's position vector r⃗\vec{r} at t=15t = 15 s, we need to substitute t=15t = 15 into the given position equations: x=−0.31(15)2+7.2(15)+28x = -0.31(15)^2 + 7.2(15) + 28 y=0.22(15)2−9.1(15)+30y = 0.22(15)^2 - 9.1(15) + 30 Calculating these: x=−0.31(225)+108+28=−69.75+108+28=66.25 mx = -0.31(225) + 108 + 28 = -69.75 + 108 + 28 = 66.25 \, \text{m} y=0.22(225)−136.5+30=49.5−136.5+30=−57 my = 0.22(225) - 136.5 + 30 = 49.5 - 136.5 + 30 = -57 \, \text{m} Thus, the position vector r⃗\vec{r} in unit vector notation is: r⃗=66.25i^−57j^ m\vec{r} = 66.25 \hat{i} - 57 \hat{j} \, \text{m} To find the magnitude and angle: ∣r⃗∣=(66.25)2+(−57)2=4389.0625+3249=7638.0625≈87.38 m|\vec{r}| = \sqrt{(66.25)^2 + (-57)^2} = \sqrt{4389.0625 + 3249} = \sqrt{7638.0625} \approx 87.38 \, \text{m} θ=tan⁡−1(−5766.25)≈−40.6∘\theta = \tan^{-1}\left(\frac{-57}{66.25}\right) \approx -40.6^\circ So, the position vector in magnitude-angle notation is: 87.38 m at −40.6∘87.38 \, \text{m} \, \text{at} \, -40.6^\circ
b
To find the velocity v⃗\vec{v} at t=15t = 15 s, we need to take the derivatives of the position functions with respect to time: vx=dxdt=ddt(−0.31t2+7.2t+28)=−0.62t+7.2v_x = \frac{dx}{dt} = \frac{d}{dt}(-0.31t^2 + 7.2t + 28) = -0.62t + 7.2 vy=dydt=ddt(0.22t2−9.1t+30)=0.44t−9.1v_y = \frac{dy}{dt} = \frac{d}{dt}(0.22t^2 - 9.1t + 30) = 0.44t - 9.1 Substituting t=15t = 15: vx=−0.62(15)+7.2=−9.3+7.2=−2.1 m/sv_x = -0.62(15) + 7.2 = -9.3 + 7.2 = -2.1 \, \text{m/s} vy=0.44(15)−9.1=6.6−9.1=−2.5 m/sv_y = 0.44(15) - 9.1 = 6.6 - 9.1 = -2.5 \, \text{m/s} Thus, the velocity vector v⃗\vec{v} is: v⃗=−2.1i^−2.5j^ m/s\vec{v} = -2.1 \hat{i} - 2.5 \hat{j} \, \text{m/s}
c
To find the acceleration a⃗\vec{a} at t=15t = 15 s, we need to take the second derivatives of the position functions with respect to time: ax=d2xdt2=ddt(−0.62t+7.2)=−0.62a_x = \frac{d^2x}{dt^2} = \frac{d}{dt}(-0.62t + 7.2) = -0.62 ay=d2ydt2=ddt(0.44t−9.1)=0.44a_y = \frac{d^2y}{dt^2} = \frac{d}{dt}(0.44t - 9.1) = 0.44 Thus, the acceleration vector a⃗\vec{a} is: a⃗=−0.62i^+0.44j^ m/s2\vec{a} = -0.62 \hat{i} + 0.44 \hat{j} \, \text{m/s}^2
Answer
The rabbit's position vector at t=15t = 15 s is 66.25i^−57j^ m66.25 \hat{i} - 57 \hat{j} \, \text{m} or 87.38 m87.38 \, \text{m} at −40.6∘-40.6^\circ. The velocity vector is −2.1i^−2.5j^ m/s-2.1 \hat{i} - 2.5 \hat{j} \, \text{m/s}, and the acceleration vector is −0.62i^+0.44j^ m/s2-0.62 \hat{i} + 0.44 \hat{j} \, \text{m/s}^2.
Key Concept
Kinematics: The study of motion without considering the forces causing it. Key equations include those for displacement, velocity, and acceleration.
Explanation
By substituting the given time into the position equations, we can find the position vector. Taking the first derivative of the position functions gives us the velocity, and the second derivative gives us the acceleration.
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