2006-06-26 02:35:21 · 9 answers · asked by maureen029 1 in Science & Mathematics ➔ Physics
330m
2006-06-26 02:40:50 · answer #1 · answered by amandeep s 3 · 0⤊ 0⤋
The displacement is given by the equation
displ= int vel*time +[.5 accl *time^2]
initial vel =40 m/s
accel = -5m/s
time = 6 sec
Hence displacement in 6 sec = 40*6 -.5*6^2 = 240-180=60 meters
2006-06-26 09:51:12 · answer #2 · answered by bashah1939 4 · 0⤊ 0⤋
X=1/2 aT^2 +VoT
X= 1/2 (-5 m/s^2) (6s)^2 + 40 m/s (6s)
X = .5 (-5) (36) m + 240 m
X = 150 m
It looks like the car would travel 150 m in 6 seconds.
So, how fast would it be going then? Carefully consider the equation V=Vo + aT
2006-06-26 09:44:44 · answer #3 · answered by enginerd 6 · 0⤊ 0⤋
d = d(0) + v(0)t + 1/2 at^2
d = 0 + 40 m/s t + 1/2 (-5) m/s/s t*t = 40*6 - 1/2 * 5 *36
= 240 - 90 = 150 meters
2006-06-26 09:40:50 · answer #4 · answered by tbolling2 4 · 0⤊ 0⤋
using the formula s=ut+1/2 at^2 we get
s=?,u=40m/s,acceleration[ a] = -5m/s2 {since its speed is decreasing},t=6
=> we get s=40x6 + 1/2 (-5)t2
=240-90
=150m
.
. . distance = 150 m
2006-06-26 09:45:47 · answer #5 · answered by expecto_patronum_accio 1 · 0⤊ 0⤋
v=u+at where v =velocity u= initial speed a=acceleration t=time in sec.
v=40+(-5)6
v=40-30
final velocity = 10
2006-06-26 09:41:56 · answer #6 · answered by masterchief irl 2 · 0⤊ 0⤋
150m.
The equation is:-
d=ut + (1/2).a.t^2
Where u= velocity=40m/s
t= time =6sec.
a=acceleration/retardation= -5m/s^2
2006-06-26 09:50:44 · answer #7 · answered by peekejee 2 · 0⤊ 0⤋
u-40m/s
a-5m/s2
t-6s
s
240+5/2. 36
240+90
330m
2006-06-26 09:44:55 · answer #8 · answered by sak007 1 · 0⤊ 0⤋
i can help you to know how to find the answer
( use the the equations of motion )
2006-06-26 09:40:01 · answer #9 · answered by Torch 3 · 0⤊ 0⤋