The answer must be in m/s squared.
2006-09-15 07:22:32 · 7 answers · asked by turnerresident 1 in Science & Mathematics ➔ Physics
If you assume that the acceleration is constant:
(3010000-5200)/2 = 1502400 m/s= average speed.
At this speed, it would take the object .0349/1502400=2.32x10^(-8) seconds to travel the give distance. Since acceleration is change in velocity divided by change in time, the acceleration is equal to 3004800/2.32x10^(-8) =1.29x10^(14) m/s^2.
2006-09-15 07:33:37 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
a =rate of change in velocity =acceleration.
change in velocity = 3010000 - 5200 = 3 004 800 M/S
How much time did it take to go .0349 m?
Avg. velocity = V END- Vstart / 2 = 1 502 400 m/s
t = s(distance)/Vavg.
t = .0349/105400 = 3.311195 Ã 10-7
a = change in velocity / t
a = 3004800 / 3.70018975 Ã 10-7
a = 9074668194842.41m/sec^2
2006-09-15 15:39:27 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Using the third equation of motion.
v^2=u^2+2as
v represents the final velocity
u represents the initial velocity
a represents the acceleration
s represents the distance
so: a=(v^2-u^2)/2s
a=(30100^2-5200^2)/(2*0,0394)
a=approximately 1,115444162*10^10 m/s^2
hope this was of some help to you
2006-09-15 14:40:59 · answer #3 · answered by Sage 1 · 0⤊ 0⤋
a = change in velocity / time
change in velocity = 3010000 - 5200 = 3 004 800
now how much time did it take to go .0349 m?
avg. velocity = end v - start v / 2 = 1 502 400
t = s/v
t = .039/105400 = 3.70018975 Ã 10-7
a = delta v / t
a = 3004800 / 3.70018975 Ã 10-7
a = .0812066
don't forget to use units!
don't forget to pay attention to significant digits!
2006-09-15 14:28:52 · answer #4 · answered by DanE 7 · 0⤊ 0⤋
You have several different answers. I'll try to break the tie. I like the formula that Sage used, discussion was good but 2 zeros were dropped from the new velocity. james k did it the hard way but got the right answer. Actually I got 1.2980047 x 10^14 m/s^2.
2006-09-15 18:37:14 · answer #5 · answered by sojsail 7 · 0⤊ 0⤋
One needs to check whether the velocity is large enough that relativistic effects need be considered. It is not, so the usual Newtonian mechanical formulas can be used. Presumably the object is an atomic particle of some kind.
2006-09-15 14:32:09 · answer #6 · answered by Anonymous · 0⤊ 0⤋
86,097,421.2 m/s^2?
2006-09-15 14:25:37 · answer #7 · answered by Anonymous · 0⤊ 0⤋