A box starts from rest at point K and travels along the horizontal conveyor. During the motion, the increase in speed is a_{t} = 0.5t m/s^{2}, where t is in seconds. Determine the magnitude of its accleration when it arrives at Q. 1) 11.25 m/s^{2 }2) 10.51 m/s^{2 }3) 12.35 m/s^{2 }4) 15.31 m/s^{2} 

The position of the box at any instant is defined from the fixed point K using the position coordinate s. The accleration is to be determined at Q, so the origin of the n,t axes is at this point. The accleration is calculated from its components a_{t} = dv/dt and a_{n}= v^{2}/r. To determine these it is first necessary to formulate v and dv/dt. Since v_{K} = 0 when t = 0, So, the accleration a_{Q}= 10.51 m/s^{2} 