An abstraction often used in physics is motion with constant acceleration. This is a good approximation for many different situations: free fall over small distances or in low-density atmospheres, full braking in car traffic, an object sliding down an inclined plane, etc … The mathematics behind this special case is relatively simple. Assume the object that is subject to the constant acceleration a (in m/s²) initially has a velocity v(0) (in m/s). Since the velocity is the integral of the acceleration function, the object’s velocity after time t (in s) is simply:
1) v(t) = v(0) + a · t
For example, if a car initially goes v(0) = 20 m/s and brakes with a constant a = -10 m/s², which is a realistic value for asphalt, its velocity after a time t is:
v(t) = 20 – 10 · t
After t = 1 second, the car’s speed has decreased to v(1) = 20 – 10 · 1 = 10 m/s and after t = 2 seconds the car has come to a halt: v(2) = 20 – 10 · 2 = 0 m/s. As you can see, it’s all pretty straight-forward. Note that the negative acceleration (also called deceleration) has led the velocity to decrease over time. In a similar manner, a positive acceleration will cause the speed to go up. You can read more on acceleration in this blog post.
What about the distance x (in m) the object covers? We have to integrate the velocity function to find the appropriate formula. The covered distance after time t is:
2) x(t) = v(0) · t + 0.5 · a · t²
While that looks a lot more complicated, it is really just as straight-forward. Let’s go back to the car that initially has a speed of v(0) = 20 m/s and brakes with a constant a = -10 m/s². In this case the above formula becomes:
x(t) = 20 · t – 0.5 · 10 · t²
After t = 1 second, the car has traveled x(1) = 20 · 1 – 0.5 · 10 · 1² = 15 meters. By the time it comes to a halt at t = 2 seconds, it moved x(2) = 20 · 2 – 0.5 · 10 · 2² = 20 meters. Note that we don’t have to use the time as a variable. There’s a way to eliminate it. We could solve equation 1) for t and insert the resulting expression into equation 2). This leads to a formula connecting the velocity v and distance x.
Solved for x it looks like this:
It’s a very useful formula that you should keep in mind. Suppose a tram accelerates at a constant a = 1.3 m/s², which is also a realistic value, from rest (v(0) = 0 m/s). What distance does it need to go to full speed v = 10 m/s? Using equation 3)’ we can easily calculate this:
Here are a few exercises and solutions using the equations 1), 2) and 3).
1. During free fall (air resistance neglected) an object accelerates with about a = 10 m/s. Suppose the object is dropped, that is, it is initially at rest (v(0) = 0 m/s).
a) What is its speed after t = 3 seconds?
b) What distance has it traveled after t = 3 seconds?
c) Suppose we drop the object from a tower that is x = 20 meters tall. At what speed will it impact the ground?
d) How long does the drop take?
Hint: in exercise d) solve equation 1) for t and insert the result from c)
2. During the reentry of space crafts accelerations can be as high as a = -70 m/s². Suppose the space craft initially moves with v(0) = 6000 m/s.
a) What’s the speed and covered distance after t = 10 seconds?
b) How long will it take the space craft to half its initial velocity?
c) What distance will it travel during this time?
3. An investigator arrives at the scene of a car crash. From the skid marks he deduces that it took the car a distance x = 55 meters to come to a halt. Assume full braking (a = -10 m/s²). Was the car initially above the speed limit of 30 m/s?
Solutions to the exercises:
a) 30 m/s
b) 45 m
c) 20 m/s
d) 2 s
a) 5,300 m/s and 56,500 m
b) 42.9 s (rounded)
c) 192,860 m (rounded)
Yes (he was initially going 33.2 m/s)
To learn the basic math you need to succeed in physics, check out the e-book “Algebra – The Very Basics”. For an informal introduction to physics, check out the e-book “Physics! In Quantities and Examples”. Both are available at low prices and exclusively for Kindle.