Question 1 A fun seeker begins sliding from the top of a water
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Question 1 A fun seeker begins sliding from the top of a water slide and accelerates uniformly at 0.86 m. s-2. If they reach a velocity of 7.78 m.s-1at the bottom of the slide, what length must the slide be?If rounding is required, please express your answerto two decimal places. Solution a=0.86,v=7.78,u=0 From equations of uniform linear motion v2=u2+2as Substituting the variables, 7.782=02+(2×0.86×s) lengthofslides=60.5284 1.72=35.1909 Question 2 If a rugby player is running at 2.82 m/s and then decides to sprint in a straight line for a ball and 2.38 seconds later is running at 7.8 m/s, what is their average acceleration during that time? Solution From linear motion equations, a=v−u t=7.8−2.82 2.38=2.0924 Question 3 A motorcycle racer falls off their motorbike while traveling at 112 km/hr and skids in a straight line down the track. If the friction with the track decelerates the rider at a uniform rate of 4.4 m/s/s, how far (what displacement) would the rider travel before coming to a stop? Solution v=0,u=112km hr(convertingspeedintom/s)=112×1000 60×60=31.111m/s a=−4ms−2 From equations of linear motion, v2=u2+2as o=31.1112+(2×−4×s) s=−967.90 −8=120.9877
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Question 4 An out of control car traveling at 60 km/hr locks its brakes and skids in a straight line. If the friction caused by locking the brakes decelerates the car at a uniform rate of 2 m/s/s, how long (what time) will it take for the car to come to rest? Solution u=60km hr=60×1000 60×60=16.6667 v=0 a=−2 From linear motion question, v=u+at t=v−u a=a=0−16.6667 −2=81 2 Question 5 A high-diver performs a handstand dive (falls vertically) from a 38 m platform. What velocity will the diver hit the water with if they fall from this height? Tip: Remember that any motion downwards is negative (-ve). Please indicate the direction of this velocity accordingly in your answer for this question. Solution a=g=9.80 u=o v=? s=38m v2=u2+2as v=√u2+2as=√2∗9.8∗38=−27.29