JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 11)
An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at $$x=+2~\mathrm{m}$$, its velocity is $$\mathrm{ - 4\widehat j}$$ m/s. The object's velocity (v) and acceleration (a) at $$x=-2~\mathrm{m}$$ will be
$$v=4\mathrm{\widehat i~m/s},a=8\mathrm{\widehat j~m/s^2}$$
$$v=4\mathrm{\widehat j~m/s},a=8\mathrm{\widehat i~m/s^2}$$
$$v=-4\mathrm{\widehat i~m/s},a=-8\mathrm{\widehat j~m/s^2}$$
$$v=-4\mathrm{\widehat j~m/s},a=8\mathrm{\widehat i~m/s^2}$$
Explanation
_29th_January_Evening_Shift_en_11_1.png)
$$\overrightarrow v = 4\widehat j$$ (m/s)
$$a = {{{v^2}} \over R} = {{16} \over 2} = 8$$ m/s$$^2$$
$$\overrightarrow a = 8\left( {m/{s^2}} \right)\left( {\widehat i} \right)$$
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