# Negative value when calculating the distance

by Anna55
Tags: distance, negative
 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!
 HW Helper Thanks P: 9,290 The acceleration is negative, a=-µg. ehild
 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
HW Helper
Thanks
P: 9,290

## 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
 P: 20 Thank you very much ehild! I understand now.

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