1. accleration in meters per secound per second.
2.distance traveled during the 12 seconds.
2007-04-20 07:21:22 · 5 answers · asked by markusa_j 1 in Science & Mathematics ➔ Mathematics
First get your units consistent.
1 km/h = 1/3.6 m/s
30 km/h = 30/3.6 m/s = u
81 km/h = 22.5 m/s = v
a = (v-u)/t = 1.18 m/s^2
s = u*t + 1/2*a*t^2
= 185 m
2007-04-20 07:27:57 · answer #1 · answered by Dr D 7 · 1⤊ 0⤋
Acceleration = v - u / t in appropriate units
u = 30 km/hr = 30,000 m/ hr = 30,000 / 3600 = 300 / 36 = 100/12 = 8.333 m/sec
v = 81 km/hr = 81000 / 3600 = 810/36 = 270 / 12 = 90 / 4 = 45 / 2 = 22.5 m/sec
v = u + at is the formula and 22.5 = 8.333 + a.12 or a = 22.5 - 8.333 / 12 = 1.18 m/sec/sec
s = v^2 - u^2 / 2a
= 22.5^2 - 8.333^2 / 2.36 = 185 m approx
2007-04-20 14:41:12 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
Average acceleration =1/12(81-30)*1000/(3600)=
51/12*5/18=1.18m/s^2
If we suppose that the acceleration is constant
velocity =1.18 t +30*1000/3600 =1.18t+8.33 m/s
and space= 1.18/2 t^2 +8.33t
At t = 12 s
space = 184.92m
2007-04-20 14:36:47 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
30000/3600 = 8.33m/s
81000/3600 = 22.5 m/s
(22.5-8.33)/12 = 1.18 m/sec^2 = acceleration
dv/dt = 1.18
v= 1.18t+c
v= 8.33 when t= 0 so c = 8.33
v= 1.18t +8.33
ds/dt = 1.18t+8.33
s = .59t^2 +8.33t +c
s=.59t^2 + 8.33t = .59(12)^2 +8.33(12) = 184.9 m
2007-04-20 14:40:06 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
1. 4.25 m/s^2
2. 666m
2007-04-20 14:25:58 · answer #5 · answered by David 3 · 0⤊ 3⤋