A particle moving along the x axis is located at 14.9m at 1.45s and 4.02m at 2.9s. What is the displacement during this time interval? What is the average velocity during this period? What would the average speed be?
Thanks in advance
2007-09-12 13:55:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The displacement = 4.02m - 14.9m = - 10.88m
The negative indicates the direction
The time = 2.9 - 1.45 = 1.45s
Average velocity = -10.88/1.45 = - 7.50 m/s
The negative indicates the direction
The average speed = 7.50 m/s
The speed is the magnitude of the velocity, direction not included
2007-09-12 14:01:50 · answer #1 · answered by Marvin 4 · 0⤊ 0⤋
Displacement along the x- axis: 4.02 - 14.9 = - 10.88 m
The formula is x2 - x1 and the negative sign indicates that the distance from the x-axis is decreasing. By convention, the motion is +ve if increasing along the x-axis (going to your right) and -ve if going to your left side.
Average velocity = Displacement / time
= - 10.88 / 1.45
= - 7. 503... m/s
Average speed = 7.5 m/s ignoring the - ve sign.
2007-09-12 21:04:10 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
t=2.9-1.45
= 1.45
d= 4.02-14.9
= -10.88
v = d/t
= 10.88 / 1.45
= 7.5 m/s
Average Speed = 7.5 m/s
Average Velocity = -7.5m/s
* minus sign in velocity
* answer must in only in one decimal place.
2007-09-12 21:05:06 · answer #3 · answered by Big_Boss 1 · 0⤊ 0⤋