An object is initially at rest moved & attains a uniform acceleration covering a distance of 7.4m in 2.5 sec. What is the acceleration of the object?
2007-10-07 04:05:09 · 4 answers · asked by Mumu [Neji] XD 1 in Science & Mathematics ➔ Physics
with solutions please..^_^
2007-10-07 04:12:51 · update #1
There are a few steps needed to solve this problem. First you need to find the object's final velocity. To do that you would need to solve distance/time.
7.4m/2.5s
Once you have that answer substitute it in the acceleration formula for final velocity. The initial velocity would be 0m/s since the object was at rest.
acceleration = (final velocity-initial velocity)/time
acceleration = ((7.4m/2.5s)- 0m/s)/ 2.5s
2007-10-07 04:14:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
the distance made by the object is
d= 1/2 *a*t^2
so 7.4 = 1/2 *a*(2.5)^2= 6.25*0.5*a
and a= 7.4/3.125=2.368m/s^2
2007-10-07 11:24:02 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
x=1/2 * a * t^2
7.4= 1/2 * a * 6,25
14,8 / 6,25 = a
a = 2,368 m/sec^2
not so much of a "normal" number...
2007-10-07 11:13:37 · answer #3 · answered by Santa's little helper 2 · 0⤊ 0⤋
7.4/2.5 that is ur acceleration... do the division.. i cant find me calculator.
2007-10-07 11:10:49 · answer #4 · answered by GalNextDoor 4 · 0⤊ 0⤋