A curvilinear motion of a particle is defined by VX= 25 - 8t m/s and y=48 - 3t2m. At t=0, x=0, Find out position, velocity and acceleration at t=4 sec.

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written 3.9 years ago by

teamques10 ★ 69k

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engineering physics

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written 3.9 years ago by

teamques10 ★ 69k

$V_x=25-8t\[2ex] \therefore x=25t-4t^2$ and ax=-8 at t=4 $x=25(4)-4(4)^2=36m\[2ex] V_x=25-8(4)=-7m/s$ $y=48-3t^2\[2ex] V_y=-6t\[2ex] a_y=-6$ At t=4 $y=0\[2ex] V_y=-24m/s$ $\therefore V_4=25m/s\swarrow74^\circ\[2ex] a_4=\sqrt{8^2+6^2}=10m/s^2\swarrow37^\circ$

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