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Unformatted text preview: Problem Set 3: Problem 2.
Problem: A man driving his car at a speed vo slams on his brakes and comes to a stop after skidding a distance xf . During the skid, his acceleration, a, is constant and the coefficient of sliding friction between the car's tires and the road is . (a) Solve for a as a function of vo and xf . (b) Determine the time it takes for the car to stop, tf , as a function of xf and vo . (c) Determine as a function of a and gravitational acceleration, g. (d) Compute the values of a, tf and for vo = 72 km/hr and xf = 40 m. Solution: The overall approach to this problem is as follows. Because we are given velocity and position data, it is most convenient to use a = vdv/dx in Part (a). Then, to determine the time for the car to stop, we use a = dv/dt in Part (b). In Part (c), we balance forces in order to determine the coefficient . Finally, in Part (d), we substitute the given values into the formulae developed in Parts (a) to (c). (a) Using the fact that vdv/dx = a and that a is constant, we have d 1 2 v 2 = adx = 1 2 1 2 v = vo + ax 2 2 where we use the fact that v = vo at t = 0 to determine the constant of integration. Now, when the car comes to a stop, v = 0. Thus, we conclude that 1 2 axf =  vo 2 Solving for a, there follows a=
2 vo 2xf The fact that a is negative tells us that the car decelerates, which is physically correct. (b) Now, we use the fact that dv/dt = a, wherefore dv = adt = v = vo + at where we again use the fact that v = vo when t = 0. Using the result for a derived in Part (a), v = vo  When v = 0, the time is t = tf , so that
2 vo tf = vo 2xf 2 vo t 2xf Solving for tf yields tf = 2xf vo (c) As shown in the figure, there are three forces acting on the car, viz., its weight W = mg, the reaction force from the road, N , and the friction force, N .
. . . . . . . . . . ... ... . .. .. . ..................... ............. . . . . . . . . . . .... ................ . ......................................... .... .. ........ . . .. .......................... ... .. .. .................................................................... ....................................................................... .. .. ...... ... . .................. . . . ...................... ... .. ........................................................ . .. ........ .......................................................... . ....................................................................................................................................................................................... . ................................................................................................................................................................................................................................................................................. . . .................................................................................................................................................................................................................. .. .... ... . .. . . . . . . . . . . W N Summing forces in the vertical direction, we find N Fz = N  W = 0 = N = W = mg The only force acting in the horizontal direction is the slidingfriction force, wherefore N = ma Hence, substituting for N yields mg = ma = = a g (d) We are given vo = 72 km/hr and xf = 40 m. First, we must convert the speed to m/sec. Hence, vo = 72 km hr 1000 m km 1 hr 3600 sec = 20 m sec Thus, the quantities a, tf and are a=
2 (20 m/sec)2 vo m = = 5 2xf 2(40 m) sec2 tf = 2xf 2(40 m) = = 4 sec vo (20 m/sec) a 5 m/sec2 = 0.51 = g 9.807 m/sec2 = ...
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 '06
 Shiflett

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