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... a toddler crawls towards a mirror at a rate of 0.25 m/s, then at what

**speed**will the toddler and the toddler's image approach each other? Answer = 0.50...4 Answers · Science & Mathematics · 13/04/2008

...total time of T=10.4s.

**Laura**accelerates for t1...then runs at constant**speed**for t2=T-t1=10.4-1...when Healan's**speed**equalled**Laura**'s.**Laura**'...2 Answers · Science & Mathematics · 09/09/2012

KE becomes friction work: ½mv² = µmgd → mass cancels d = v² / 2µg = (6.0m/s)² / (2*0.12*9.8m/s²) = 1.5 m If you find this helpful, please award Best Answer. You get points too!

2 Answers · Science & Mathematics · 13/11/2017

**Laura**could not view Lily waving before unless Lily was moving faster than light...local measurement, not fixed. So, I would say yes to the question "can everyone have there [sic] own time**speed**?." Ask any 3 people what time they have, and you will get 3 different answers, assuming they ...3 Answers · Science & Mathematics · 13/05/2007

...top

**speed**10,4 - 2 = 8,4 s Healen race at top**speed**10,4 - 3 = 7,4 s**Laura**top**speed**= aL * t acceleration = aL * 2 Healen top**speed**= aH * t...2 Answers · Science & Mathematics · 15/02/2012

.... (accelerating and maximum

**speed**)**Laura**: During beginning we know that...1 * (6-3) + 0 s = 40.5m Therefore,**Laura**is ahead assume they started at the...1 Answers · Education & Reference · 31/01/2013

First deal with

**Laura**: She ran 10.4-2.0 = 8.4 sec running at constant**speed**Call this constant**speed**... Final answer:**Laura**'s acceleration = 5.32m/s ...2 Answers · Science & Mathematics · 13/09/2010

...original race had 100 yds. Lets set an example that this took

**Laura**10 seconds. So her average**speed**was 10 yds/sec. When**Laura**is done...5 Answers · Science & Mathematics · 16/10/2007

...100 = (0.5)(2.17)L + 8.23L ==> L = 10.735 m/sec max

**speed**for**Laura**Likewise: 100 = (0.5)(3.24)H + 7.16H ==> H...1 Answers · Science & Mathematics · 27/12/2011

...she pull over and get a bite to eat. *Now that you've provided

**Laura**'s**speed**, Mark would pass her at just after 5:00pm**Laura**2 hrs...3 Answers · Science & Mathematics · 12/02/2009

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