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At the instant shown, cars A and B travel at speeds of 70 mi / h and 50 mi / h, respectively. If B is increasing its speed by 1100mi/h21100 mi / h ^2, while A maintains a constant speed, determine the velocity and acceleration of B with respect to A. Car B moves along a curve having a radius of curvature of 0.7 mi.


Answered 2 years ago
Answered 2 years ago
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Velocity.\text{\color{#c34632}Velocity.} The origin of the xx and yy axes are located at an arbitrary fixed point. Since the motion relative to car AA is to be determined, the translating frame of reference x, y{x}',~{y}' is attached to it. Applying the relative-velocity equation, we have

vB=vA+vB/A\bold{v}_B = \bold{v}_A + \bold{v}_{B/A}

The direction of motion of car BB is at 3030^\circ with the vertical. Realizing that vB=50v_B = 50 km/h and vA={70j}\bold{v}_A = \{70\bold{j}\} km/h, then

50sin(30)i+50cos(30)j=70j+vB/A50\sin(30^\circ)\bold{i} + 50\cos(30^\circ)\bold{j} = 70\bold{j} + \bold{v}_{B/A}

vB/A={25i26.7j} m/s\bold{v}_{B/A} = \{25\bold{i} - 26.7\bold{j}\}\mathrm{~m/s}

So that,

vB/A=(25)2+(26.7)2=36.58 km/hv_{B/A} = \sqrt{(25)^2 + (-26.7)^2} = \boxed{\color{#4257b2}36.58\mathrm{~km/h}}

Noting that vB/A\bold{v}_{B/A} has +i+\bold{i} and j-\bold{j} components, as shown in the Fig., its direction is therefore

tanθ=(vB/A)y(vB/A)x=26.725\tan\theta = \frac{(v_{B/A})_y}{(v_{B/A})_x} = \frac{26.7}{25}

θ=46.88   d\boxed{\color{#4257b2}\theta = 46.88^\circ~~~d}

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