Question #241986

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by 𝑣 = [5.00π‘š/𝑠 βˆ’ (0.0180π‘š/ 𝑠 3 )𝑑 2 ]𝑖̂+ [2.00 π‘šβ„π‘  + (0.550 π‘š 𝑠 2 ⁄ )𝑑]𝑗. (a) What are Μ‚ π‘Žπ‘₯(𝑑) and π‘Žπ‘¦(𝑑), the π‘₯- and 𝑦-components of the car’s velocity as functions of time? (b) What are the magnitude and direction of the car’s velocity at 𝑑 = 8.00 𝑠? (b) What are the magnitude and direction of the car’s acceleration at 𝑑 = 8.00 𝑠?


1
Expert's answer
2021-09-29T09:55:51-0400

(a) The x- and y-component of car's velocity:


vy=[2.00+(0.550)𝑑] m/s,vx=[5.00βˆ’(0.0180)𝑑2] m/s.v_y=[2.00+ (0.550 )𝑑] \text{ m/s},\\ v_x=[5.00 βˆ’ (0.0180)𝑑^2 ]\text{ m/s}.


Find the acceleration:


a(t)=vβ€²(t)=[βˆ’0.036t]i^+[0.55]j^.ax(t)=βˆ’0.036t,ay(t)=const=0.55 m/s2.a(t)=v'(t)=[-0.036t]\hat i+[0.55]\hat j.\\ a_x(t)=-0.036t,\\ a_y(t)=\text{const} =0.55\text{ m/s}^2.

(b) The magnitude:


v=(2+0.55β‹…8)2+(5βˆ’0.018β‹…82)2=7.47m/s, ΞΈ=arctan⁑2+0.55β‹…85βˆ’0.018β‹…82=58.0Β°v=\sqrt{(2+0.55Β·8)^2+(5-0.018Β·8^2)^2}=7.47\text{m/s},\\\space\\ \theta=\arctan\frac{2+0.55Β·8}{5-0.018Β·8^2} =58.0Β°

from +i toward +j.


a=(βˆ’0.036β‹…8)2+(0.55)2=0.620 m/s2. Ο•=arctan⁑0.55βˆ’0.036β‹…8=βˆ’62.4Β°a=\sqrt{(-0.036Β·8)^2+(0.55)^2}=0.620\text{ m/s}^2.\\\space\\ \phi=\arctan\frac{0.55}{-0.036Β·8}=-62.4Β°

from +i toward +j (or 62.4Β° from +i to +j). 


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