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acceleration is the change in velocity over time. so what are the units of acceleration?

2006-09-28 15:16:57 · 7 answers · asked by foxy lady 2 in Science & Mathematics Physics

7 answers

Acceleration is expressed as a distance unit divided by a time unit squared, or more unusally, a force divided by a mass.

Usual, metric (MKS) is meters/second^2 or m/s^2.

Unusual, but encountered, N/kg (which reduces to m/s^2). Used to emphasize the effect of gravity fields and inertia.

2006-09-28 15:28:26 · answer #1 · answered by Anonymous · 0 1

acceleration is the rate of change of speed.

speed is measured in units of distance per time.

acceleration is then the rate of distance per time per unit time.

this works out distance per time squared.

for example, if a car changes in speed from 10 meters/sec to 20 meters/sec in 1 second, then its acceleration is 10 meters/sec/sec or 10 meters/ sec^2 ( squared ).

acceleration can also be measured in "g's" ( e.g "pulling g's" ).

1 g is the force experienced by the gravitational pull of the earth ( measured as 9.8 meters/sec/sec )

thus, if you are in car and experiencing 1g , then the car is presumed to accelerating at 9.8 meters/sec/sec.

this is just a convenient way of expressing acceleration.

2006-09-28 16:36:04 · answer #2 · answered by fullbony 4 · 1 0

F=ma so a = F/m

But usually we think of a as the rate of change of velocity dv/dt,which is (in mks) (m/s)/s or m/s^2

distance per time^2

2006-09-28 15:37:40 · answer #3 · answered by bubsir 4 · 0 0

Usually in school they are using the metric system.

Units of acceleration are meters/second^2
or meters per second squared.

Gravity is 9.8meters/second squared.

2006-09-28 15:23:47 · answer #4 · answered by Rich S 1 · 0 0

ft per second^2

2006-09-28 15:18:46 · answer #5 · answered by Anonymous · 0 0

any unit of distance over any unit of time

2006-09-28 15:20:53 · answer #6 · answered by ay3e 2 · 0 2

meters per second square

feet per second square


I guess... miles per hour square

2006-09-28 15:24:07 · answer #7 · answered by J G 4 · 0 0

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