Negative value when calculating the distance


by Anna55
Tags: distance, negative
Anna55
Anna55 is offline
#1
Oct23-11, 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=g
2 g Δx=vf2- vo2
vf=0
2 g Δx=- vo2
Δx=- vo2/2 g
Δx=-252/(20.959.8)
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!
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ehild
ehild is offline
#2
Oct23-11, 02:16 AM
HW Helper
Thanks
P: 9,818
The acceleration is negative, a=-g.

ehild
Anna55
Anna55 is offline
#3
Oct23-11, 06:21 AM
P: 20
If I have understood this correctly, the solution should be this:

2aΔx=vf2- vo2
a=-g
2 -g Δx=vf2- vo2
vf=0
2 -g Δx=- vo2
Δx=- vo2/2 -g
Δx=-252/(2-0.959.8)
Δx=33.6 m

ehild
ehild is offline
#4
Oct23-11, 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 -g Δx means subtraction. Correctly: 2(-g)Δx or -2g Δx. And 25^2=625.

Written correctly your derivation, it looks :

2aΔx=vf^2- vo^2
a=-g
2 (-g) Δx=vf^2- vo^2
vf=0
2 (-g)Δx=- vo^2
Δx=- vo^2/(-2g)
Δx=-625/(2(-0.95)9.8)
Δx=33.6 m

ehild
Anna55
Anna55 is offline
#5
Oct23-11, 10:38 AM
P: 20
Thank you very much ehild! I understand now.


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