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Hey mates I will mark u as brainliest plz answer this... A cyclist is riding with a speed of 27 km/h. As heapproaches a circular turn on the road of radius 80 m, heapplies brakes and reduces his speed at the constant rate of0.5 m/s2. What is the magnitude and direction of the netacceleration of the cyclist on the circular turn? |
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Answer» tion:0.86 m/s2; 54.46° with the direction of velocitySpeed of the cyclist, v = 27 km/h = 7.5 m/sRadius of the CIRCULAR turn, r = 80 mCentripetal acceleration is given as:ac = v2 / r= (7.5)2 / 80 = 0.7 ms-2Suppose the cyclist begins CYCLING from POINT P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.This acceleration is ALONG the TANGENT at Q and opposite to the direction of motion of the cyclist.Since the angle between ac and aT is 900, the resultant acceleration a is given by:a = (ac2 + aT2)1/2= ( (0.7)2 + (0.5)2 )1/2= (0.74)1/2 = 0.86 ms-2tan θ = ac / aTwhere θ is the angle of the resultant with the direction of velocity.tan θ = 0.7 / 0.5 = 1.4θ = tan-1 (1.4) = 54.560 hope my answer helped you!plz mark this answer as brainliest! |
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