I know that the rate of acceleration for a falling object on earth (barring air resistance) is 9.8 meters per second squared. Some sources state this as “per second per second”. This is what I am having trouble understanding. How can a unit as such time be squared? When applying this equation to, say, dropping a rock down into the Grand Canyon, how would I determine how fast the rock is falling after x number of seconds? If x is 3, which when squared is 9, how does this apply to the time that has elapsed? I know from my own experiments that if I drop an object it hasn’t traveled 9.8 meters after one second (one squared equals one, of course), so what am I missing here. How would I use this equation to determine how fast the rock is moving after ten seconds have elapsed? Lastly, how I would I use this equation to determine the rate of acceleration for a car that goes from zero to 100 kmph in ten seconds? (I’m using metric units for ease of calculation). How fast would the car be going if it continued to accelerate at this rate at 20 seconds?
Also, I have a vague idea of what terminal velocity is, but can someone explain this to me and how it is determined?
Maybe it’s easier to think about it this way: not 9.8 meters per second squared, but 9.8 meters per second per second: that is, the velocity of your rock will increase by 9.8 meters per second each second.
I think that these are the formulas you need:
v=a*t
v is velocity, a is the acceleration, and t is time. You can use this to find the velocity of your rock at any given time, and you can use a=v/t to find the average acceleration of the car (10 km per hour per second, or 2.8 meters per second squared.)
x=1/2 a t^2
x is the distance traveled.
To find out how far a rock has traveled in 1 second, you square the second (1) and multiply by the acceleration and divide by 2, and you get 4.9 m. (You’re probably wondering where the 1/2 comes from, and the answer is that it’s calculus, which isn’t a very satisfying explanation, I’ll grant you . . .)
Terminal velocity is the speed at which the force of air friction on an object equals the force due to gravity. The two forces cancel out and the object just falls at a constant velocity. The force due to air friction depends on the density of the air, surface area presented by the object and its velocity, and this whacky thing called the drag coefficient that takes into a account a lot of things (slipperiness, turbulence, etc) that are difficult or impossible to compute mathematically.
You’re making it harder than it is. m/s[sup]2[/sup] is just another way of saying m/s/s.
Your care example is simple. If it accelerates at a constant rare to 100mph in 10 second is accelerates at 10mph per second or 17.6 feet per second pers second. At the end of 20 seconds it would be going 200mph. Of coruse all this assumes it’s in a vacuum because aerodynamic drag increases with the cube of speed. Double the speed and drag is eight times as much.
velocity = acceleration * time
To find the current velocity of an accelerating object you multiply acceleration by time. That means after one second the an object will be falling at a speed of 9.8 m/s. After two seconds it will be falling at a speed of 19.6 m/s and so on.
If an object is moving at a constant speed you can find its distance traveled by multiplying velocity by time:
position = velocity * time
But if the velocity is changing over time this doesn’t work … as you’ve discovered by your experiments. What you want is not the velocity at a particular time but the average velocity over the whole interval you’re interested it.
Because the velocity is increasing linearly you can find the average by dividing the final velocity in half.
position = average velocity * time
position = 1/2 * final velocity * time
position = 1/2 * acceleration * time * time
Is that clear?
Terminal velocity is caused by air resistance. It’s the point where the upward force created by air resistance balances the downward force created by gravity. With the forces balanced the falling body stops accelerating and reaches its maximum falling velocity.