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 300 km/h in a semicircular arc with a radius of 220 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. 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 Okay, so here we have our semicircle.

00:04 Our car is going to be going from zero to a speed of 300 kilometers per hour, which is the same as 83 .3 meters per second.

00:12 And we can find the tangential acceleration, we're told, is constant.

00:19 So we can use this whole arc to find what that is.

00:24 And the arc length will be pi times the radius, so half the circle.

00:31 Using that we can then use our time -independent kinematic equation the final speed that's 83 .3 meters per second squared is equal to the initial speed squared plus two times the acceleration times the arc the length of travel the displacement and this term right here is going to be zero so we can drop that and then solve for a so the tangential acceleration is going to equal v final squared divided by two times the arc length, which is pi r.

01:05 So the final squared over two pi r.

01:10 Plugging in our values for the velocity and the radius, we get a tangential acceleration of 5 .02 meters per second squared.

01:23 And that will be the same at every point along this arc.

01:26 So at this point, p.

01:28 The centripetal acceleration at this point, we first need to find its speed at that point.

01:34 And it will so we can use our same equation but evaluating this velocity here instead of being the final velocity it's the velocity at point p.

01:44 So velocity at point p squared is equal to the initial velocity, which again will be zero plus two times the acceleration, which we just found, times s over two.

01:56 So half that arc length.

01:59 And so if we plug in for our acceleration, solving this...

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