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 350 km/h in a semicircular arc with a radius of 220 m . a. Determine the tangential and radial acceleration of the car when it is halfway through the arc, assuming constant tangential acceleration. Express your answer using three significant figures separated by a comma. 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 The circular track is shown by the red color.

00:03 A is the starting point of the track where initial velocity is zero.

00:09 C is the end point of the track where the car attends a velocity of 350 km per hour or 97 .2 meter per second.

00:21 And b is the midpoint of the track.

00:25 B is the midpoint of the track, halfway of the track.

00:33 And x is the inner age of the track, this one, and y is the outer edge of the track.

00:41 And the radius of the track is 220 meters.

00:46 Now we need to in part a, we need to find out the tangential and radial acceleration at point b, which is the halfway.

00:55 Now at point b, we have two type of acceleration.

01:01 One is the tangential acceleration, which is tangent.

01:04 To this semicircle and other is the radial acceleration which is indicating towards the means along the radius towards the center of the circle that the radial acceleration keeps it in a circular motion so we need to find out ar and a t now first we find out it is given that the tangential acceleration is constant it is given in the problem tangential acceleration is constant.

01:44 So first we try to find tangential acceleration based on the data provided.

01:49 We will apply the equation of motion v square minus u square equal to 2 as between point a and c, between point a starting point and the endpoint c.

02:03 Right.

02:04 Now here at point a initial velocity is 0 at point b the final velocity is 350 km per hour means 97 .2 to meter per second right and the distance is nothing but the perimeter of a semicircle a to c is a semi -circle so its perimeter is a semicircle so its perimeter is full circle perimeter divided by 2 means pi r means 3 .14 and r is 220 meter.

02:42 So it is 690 .8 meter.

02:47 Right? this is the value of s.

02:53 So we use all these values and it will be 97 .22 whole square minus u is 0.

03:03 0 whole square is equal to 2.

03:06 A multiplied by s is a distance between a to b in a semicircular part.

03:12 690 .8.

03:14 So a comes out to be 6 .84 meter per second square.

03:21 And this acceleration is nothing but the tangential acceleration.

03:29 Because we are calculating it means along this path.

03:34 When particle is moving, its velocity is like this.

03:38 So it is all about tangential velocity and tangential acceleration.

03:43 So we can make out the 80 is 6 .84 meter per second square.

03:54 Now as 80 is constant given in the problem, so it means at mid point b, at mid point b, the tangential acceleration is 6 .84 meter per second square.

04:18 This is one part of the solution.

04:21 Now to find radial acceleration, we know that radial acceleration a .r is equal to v square by r where v square is the tangential velocity.

04:34 To find radial acceleration at the midpoint b, we need to find out the velocity at the midpoint b.

04:43 For this, we apply the equation of motion b square minus u square equal to 2as between a to b, not going to be...

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