The particle P starts from rest at point A at time t = 0 and changes its speed t

Question

The particle P starts from rest at point A at time t = 0 and changes its speed thereafter at a constant rate of 1.7g as it follows the horizontal path shown. Determine the magnitude and direction of its total acceleration (a) just before point B, (b) just after point B, and (c) as it passes point C. State your directions relative to the x-axis shown (CCW positive) and choose the angle with the smallest magnitude.

i figured out A) a = 16.66 because its just 1.7g = 1.7(9.8), and theta = 0 becuase its horizontal

However I am confused on where you go from here.

Explanation / Answer

find speed at B

v=(2*a*s)^0.5=(2*1.7*9.8*2.6)^0.5=9.3m/s

centripetal acceleration = v^2/r=21.66m/s2

magnitude = (At^2+Ac^2)^0.5=(16.66^2+21.66^2)^0.5=27.33m/s2

at angle arctan(21.66/16.66)=52.43 degree below the x axis

acc=16.66i-27.33j

2.)at C velocity = (2*a*s)^0.5=(2*1.7*9.8*(2.6+pi*4/2))^0.5

=17.2m/s

centripetal acceleration = v^2/r=74m/s2

magnitude = (At^2+Ac^2)^0.5=(16.66^2+74^2)^0.5=75.85

acc=-74i-16.66j

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