
#1
Oct2311, 01:39 AM

P: 20

A 1000 kg car travelling rightward at 25 m/s slams on the breaks and skids to a stop (with locked wheels). Find the distance required to stop it. Coefficient of friction between the tires and the road is 0.95. Let g=9.8 m/s^2
Solution: 2aΔx=vf2 vo2 a=µg 2 µg Δx=vf2 vo2 vf=0 2 µg Δx= vo2 Δx= vo2/2 µg Δx=252/(2×0.95×9.8) Δx=33.6 m The answer became 33.6 m. Why is the value negative? The correct answer is 33.6. Thank you in advance! 



#2
Oct2311, 02:16 AM

HW Helper
Thanks
P: 9,818

The acceleration is negative, a=µg.
ehild 



#3
Oct2311, 06:21 AM

P: 20

If I have understood this correctly, the solution should be this:
2aΔx=vf2 vo2 a=µg 2 µg Δx=vf2 vo2 vf=0 2 µg Δx= vo2 Δx= vo2/2 µg Δx=252/(2×0.95×9.8) Δx=33.6 m 



#4
Oct2311, 07:38 AM

HW Helper
Thanks
P: 9,818

Negative value when calculating the distance
To denote power, use "^". Use parentheses when multiplying with a negative number. 2 µg Δx means subtraction. Correctly: 2(µg)Δx or 2µg Δx. And 25^2=625.
Written correctly your derivation, it looks : 2aΔx=vf^2 vo^2 a=µg 2 (µg) Δx=vf^2 vo^2 vf=0 2 (µg)Δx= vo^2 Δx= vo^2/(2µg) Δx=625/(2×(0.95)×9.8) Δx=33.6 m ehild 



#5
Oct2311, 10:38 AM

P: 20

Thank you very much ehild! I understand now.



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