On a dry road, a car with good tires may be able to brake with a constant deceleration of 5.01 m/s2. The car, initially traveling at 26.3 m/s, takes 524950099 s to stop. How far does it travel in this time?
2007-03-08 19:52:36 · 3 answers · asked by Wilson J 4 in Science & Mathematics ➔ Mathematics
yer its a typo ahaha, thx for pointin it out lol. its supposed to be 5.24950099
thx for the help.
2007-03-08 20:18:23 · update #1
The appropriate equation is as follows:
d = v(i)t + 0.5at^2
where:
d is the distance that the car travels
v(i) is the initial velocity
t is the time it travels
a is the acceleration
Since the car is decelerating, it has a negative acceleration. That is a = -5.01 m/s2
(26.3 m/s)(5.25 s) + 0.5 (-5.01 m/s2)(5.25 s)^2 = 69 m
2007-03-08 20:11:00 · answer #1 · answered by Sam 5 · 0⤊ 0⤋
v² = u² - 2 a s (seems to ring a bell from years gone by?)
0 = 26.3² - 2 x 5.01 x s
s = 26.3² / 10 .02
s = 69 m
2007-03-09 04:24:17 · answer #2 · answered by Como 7 · 0⤊ 0⤋
omg this is physics.. do you have to use the equation for deceleration? im so sorry i did physics last year and i cant remember!
2007-03-09 03:57:38 · answer #3 · answered by Cheerios17 2 · 0⤊ 0⤋