## Related questions with answers

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 $1100 mi / h ^2$, 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.

Solution

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

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

The direction of motion of car $B$ is at $30^\circ$ with the vertical. Realizing that $v_B = 50$ km/h and $\bold{v}_A = \{70\bold{j}\}$ km/h, then

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

$\bold{v}_{B/A} = \{25\bold{i} - 26.7\bold{j}\}\mathrm{~m/s}$

So that,

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

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

$\tan\theta = \frac{(v_{B/A})_y}{(v_{B/A})_x} = \frac{26.7}{25}$

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

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