# Homework Help: Negative value when calculating the distance

1. Oct 23, 2011

### Anna55

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. Oct 23, 2011

### ehild

The acceleration is negative, a=-µg.

ehild

3. Oct 23, 2011

### Anna55

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. Oct 23, 2011

### ehild

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. Oct 23, 2011

### Anna55

Thank you very much ehild! I understand now.