A car at the indianapolis 500 accelerates uniformly from the...

A car at the Indianapolis 500 accelerates uniformly from the pit area, going from rest to 340 km/h in a semicircular arc with a radius of 210 m . Part A: Determine the tangential and radial acceleration of the car when it is halfway through the arc, assuming constant tangential acceleration. Express your answers using three significant figures separated by a comma. Part B If the curve were flat, what would the coefficient of static friction would be necessary between the tires and the road to provide this acceleration with no slipping or skidding? Express your answer using three significant figures.

Transcript

00:01 In this question we have given that a car at the indian police accelerates uniformly from the pit area going from rest to 340 km per hour in a semicircular arc with a radius of 210 meter.

00:15 Then we have to determine the tangential and radial acceleration of the car when it is halfway through the arc.

00:26 So first of all we have given that the initial velocity of the car, that is u.

00:36 Is 0 meter per second and we have given the final velocity that is v is 340 kilometer per hour so we have to convert it in meter per second so we can say that this will come out 94 .44 meter per second we have given the radius of that arc that is 210 meter so we can say that distance covered by car that is equal to as is equal to this is 2 pi r by 2 that is equal to 210 pi now we can say that according to third equation of motion that is v square is equal to this is u square plus 2 as so from here we get the value of the tangential acceleration so we can say that the tangential acceleration 80 is this is 94 .44 square upon 2 into s is 210 pi and this will comes out 6 .76 meter per second square this is the tangential acceleration and we can say that now since the centipater acceleration depends on the speed so it increases around turn speed at half way point now we have to find first we have to calculate the speed so we can say that according to second or we can say that third equation of motion that is v1 is square is equals to this is u square plus 2 into tangential acceleration now the displacement is given to as that is this is half way so this is s by 2 so we can say that this is 2 into 6 .7 26 into this is 210 pi by 2 so we have to calculate this velocity and this comes out 66 .78 meter per second so we can say that the centipater acceleration is given by that is v1 square by r so this is 66 .78 square by 210 and that is equal to 21 .238 meter per second square square so this is the tangential and centipital acceleration...

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