A car is traveling around a circular track with a radius of400ft. An observer at

Question

A car is traveling around a circular track with a radius of400ft. An observer at the center of the circle notes that the angleis increasing such that at the instant shown the rate of change ofthe angle is = 0.2rad/sec [ has a dot over it] ( abit less than 11.5o/sec). Furthermore that the rate hasbeen increasing and the observer estimates (with 2 dots overit) =0.02rad/sec2. Determine both the velocity andacceleration of the car at the instant shown (you may leave theanswers in terms of vector components as long as you clearly defineyour coordinate system)                         withone dot

Explanation / Answer

It is important to note that at the instant shown, centripetaldirection is -y    and tangential direction(direction car is moving) is -x if we use traditionaldirections of   up for y and right for x. (I thinkyour picture shows the car moving counterclockwise... let me knowif this is wrong. . Then the velocity, at that instant, is only tangential and isequal to r = (0.2 rad/sec ) (400 ft) =   80 ft/sec .     (note:    is thesame thing as dot, which also means d/dt) . So the velocity is       80 ft/sec   in the -y direction,   or    -80 ft/sec   y direction . The acceleration has two components. Tangentialcomponent is simply the rate of change of speed, r(d/dt) = .             =   400 ft * (0.02 rad/s2) =     8.0 ft/s2 .    the centripetal component is  v2 / r    which is also  2 r = 0.22 *400   =   16 ft/s2 .      so the total accelerationis     a   =  -8.0 i    -   16j              where i and j mean x and y directions .      so the total accelerationis     a   =  -8.0 i    -   16j              where i and j mean x and y directions

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