A 10kg object is rest a point A. The object accelerates uniformly from point A to point B in 4 seconds, attaining a maximum speed of 10 meters per second at point B.
What distance did the object travel in moving from point A to point B?
(1) 2.5 m
(2) 10m
(3) 20m
(4) 100m
Please use the common formulas and explain it!
2007-02-13 19:05:12 · 4 answers · asked by Kala J 3 in Science & Mathematics ➔ Physics
How did you know the average speed was 5m/s??
2007-02-13 19:20:19 · update #1
Hmm... let's try it this way...
The object is accellerating from 0 to 10 at a uniform rate.
Therefore, the average speed is 5.
rate x time = distance
5m/s X 4 seconds = 20 meters
*EDIT IN RESPONSE TO THE FOLLOWUP QUESTION*
The average speed is 5 because the accelleration from 0 to 10 is uniform, and because the start and end of the accelleration is the same as the start and end points of the distance traveled. This means that at no point does it travel at any one speed for a longer period of time than it's traveled at any other speed.
Therefore, we can take a simple average between 0 and 10 to find the average rate as follows:
(start rate + end rate) / 2
(0 +10) / 2 = 5
Therefore, the average speed is 5m/s
2007-02-13 19:11:59 · answer #1 · answered by DavidGC 3 · 0⤊ 0⤋
Answer 20 m
Distance travelled = Average speed x time
= 1/2 [initial speed + final speed] x 4
= 1/2 [ 0 + 10 ] x 4
Initial speed = 0 as the object started from rest.
2007-02-14 03:46:44 · answer #2 · answered by pete 2 · 0⤊ 0⤋
Well answered.
The mass of the object is irrelevant, by the way... since you already know the acceleration.
The average speed is 5 because the final speed is 10, the initial speed is 0. The average is 5. This only works for uniform acceleration.
2007-02-14 03:25:14 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Where x is position, v is velocity and t is time:
x = x0 + 1/2 (v + v0)t
Substitute the known variables:
x= 0 + 1/2 (10 +0)4
x=1/2(10)4
x=20 meters
2007-02-14 03:33:29 · answer #4 · answered by Zachary F 2 · 0⤊ 0⤋